1. Introduction
Let \(A\) denote the class of all functions, \(f(z)\) normalized by
\begin{align}\tag{1}\label{eq1.1}
f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\,,
\end{align}
which are analytic in the unit disc \(U=\lbrace z:|z|< 1\rbrace\).
Definition 3. [1]
For \(f(z)\in A\) , Opoola introduced the following operator:
\begin{align}\tag{2}\label{eq1.2}
D^{0}(\mu,\beta,t)f(z)&=f(z),\notag\\
D^{1}(\mu,\beta,t)f(z)&=zD_{t}f(z)=tzf'(z)-{z(\beta-\mu)t}+[1+(\beta-\mu-1)t]f(z),\notag\\
D^{n}(\mu,\beta,t)f(z)&=zD_{t} (D^{n-1}(\mu,\beta,t)f(z)),& n\in N.
\end{align}
If \(f(z)\) is given by (\ref{eq1.1}), then from (2), we see that
\begin{align}\tag{3}\label{eq1.3}
D^{n}(\mu,\beta,t)f(z)=z+\sum_{k=2}^{\infty}\left[1+\left(k+\beta-\mu-1\right) t\right] ^{n}a_kz^{k}\,,
\end{align}
\( (0\leq\mu \leq \beta\), \(t\ge 0\) and \(n\in N_{0}=N \cup{0})\).
Remark 1.
- When \(\beta=\mu\), \(t=1\),\( D^{n}(\mu,\beta,t)f(z)= D^{n}f(z)\) by Salagean [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:
\begin{align}\tag{4}\label{eq1.4}
f(z)=z^{p} + \sum_{k=p+1}^{\infty}a_kz^{k},\;\;\;\;\;(p=1,2,…)\,,
\end{align}
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 \(f(z)\in A_p\)
\begin{align}\tag{5}
D^{0}(\mu,\beta,t,p)f(z)& =f(z),\notag\\
D^{1}(\mu,\beta,t,p)f(z)& =zD_{t,p}f(z)=\frac{t}{p}zf'(z)-z^{p}(\beta-\mu)t+[1+(\beta-\mu-1)t]f(z),\notag\\
D^{n}(\mu,\beta,t,p)f(z)& =zD_{t,p} (D^{n-1}(\mu,\beta,t,p)f(z)), & n\in N.\label{eq1.5}
\end{align}
If \(f(z)\) is given by (\ref{eq1.4}), then from (5), we see that
\begin{align}\tag{6}\label{eq1.6}
D^{n}(\mu,\beta,t,p)f(z)=z^{p}+\sum_{k=p+1}^{\infty}\left[ 1+\left( \frac{k}{p}+\beta-\mu-1\right)t\right]^{n}a_kz^{k}\,,
\end{align}
\( (0\leq\mu \leq \beta\), \(t\ge 0\) and \(n\in N_{0}=N \cup{0})\).
Let \(T_{p}\) denote the subclass of \(A_{p}\) consisting of functions of the form
\begin{align}\tag{7}\label{eq1.7}
f(z)=z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k} ,&& (a_k\ge 0,p=1,2,…)\,.
\end{align}
If \(f(z)\) is given by Eq. (\ref{eq1.7}), then from Eq. (5), we get
\begin{align}\tag{8}\label{eq1.8}
D^{n}(\mu,\beta,t,p)f(z)=z^{p}-\sum_{k=p+1}^{\infty}\left[ 1+\left( \frac{k}{p}+\beta-\mu-1\right)t\right]^{n}a_kz^{k},
\end{align}
\((n\in N_{0},a_k\ge 0,p=1,2,..,0\leq\mu \leq \beta, t\ge 0,n\in N_{0}=N \cup{0})\,.\)
Remark 2.
When \(\beta=\mu\) in (\ref{eq1.8}),\(D^{n}(\mu,\beta,t,p)f(z)=D^{n}_{\delta,p}f(z)\) defined by
Bulut in [4]. Now, from (\ref{eq1.8}), 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 (\ref{eq1.7}), while $$g(z)=z^{p}-\sum_{k=p+1}^{\infty}\left[ 1+\left( \frac{k}{p}+\beta-\mu-1\right)t\right]^{n}z^{k}\,.$$
Definition 2.
A function \(f(z)\in T_{p}\) is in the class \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) if
\begin{align}\tag{9}\label{eq1.9}
\left|\frac{(D^{n}(\mu,\beta,t,p)f(z))’-pz^{p-1}}{\lambda (D^{n}(\mu,\beta,t,p)f(z))’+(\alpha-\gamma)}\right|< \delta, && (z\in U,n\in N_{0})\,,
\end{align}
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 (\ref{eq1.8}).
Remark 3.
When \(\mu=\beta\) in (\ref{eq1.9}),the class \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) reduces to the class \(R^{n}_{p}(\alpha,\beta,\gamma,\mu)\) studied by Bulut in [4].
Definition 4. [5, 6]
The fractional integral of order \(l\) is defined, for function \(f(z)\) by
\begin{align}\tag{10}\label{eq1.10}
D_{z}^{-l}f(z)=\frac{1}{\Gamma (l)}\int_{0}^{z}\frac{f(t)}{(z-t)^{1-l}}d(t),&&(l>0)\,,
\end{align}
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
\begin{align}\tag{11}\label{eq1.11}
D_{z}^{l}f(z)=\frac{1}{\Gamma (1-l)}\frac{d}{d(z)}\int_{0}^{z}\frac{f(t)}{(z-t)^{l}}d(t),& & (0\le l< 1)\,,
\end{align}
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
\begin{align}\tag{12}\label{eq1.12}
D_{z}^{p+l}f(z)=\frac{d^{p}}{d(z)^{p}}D_{z}^{l}f(z), & &(0\le l< 1,p\in N_{0})\,.
\end{align}
It readily follows from (\ref{eq1.9}) and (\ref{eq1.10}) that
\begin{align}\tag{13}\label{eq1.13}
D_{z}^{-l}z^{k}=\frac{\Gamma (k+1)}{\Gamma (k+l+1)}z^{k+l},& & (l>0,k\in N)\,,
\end{align}
and
\begin{align}\tag{14}\label{eq1.14}
D_{z}^{l}z^{k}=\frac{\Gamma (k+1)}{\Gamma (k-l+1)}z^{k-l}, & & (0\le l< 1,k\in N)\,.
\end{align}
Lemma 1.[7]
If \(f(z)\) and \(g(z)\) are analytic in \(U\) with \(f(z) \prec g(z)\), then for \(\sigma >0\) and \(z=re^{i\theta}\), \({(0 < r< 1)}\), then
\[\int_{0}^{2\pi}\left|f(z)\right|^{\sigma}d\theta \le \int_{0}^{2\pi}\left|g(z)\right|^{\sigma}d\theta\,.\]
In this work, several properties of the class \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) 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 \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) if and only if
\begin{align}\tag{15}\label{eq1.15}
\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k \le \delta(\lambda p+\alpha-\gamma)\,,
\end{align}
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{align*}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{align*}
Proof
Suppose that \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta) \), then we have from (\ref{eq1.9}) that
$$\left|\frac{(D^{n}(\mu,\beta,t,p)f(z))’-pz^{p-1}}{\lambda (D^{n}(\mu,\beta,t,p)f(z))’+(\alpha-\gamma)}\right|< \delta\,.$$
By substitution, we have
$$\left|\frac{pz^{p-1}-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1}-pz^{p-1}}{\lambda (pz^{p-1}-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1})+(\alpha-\gamma)}\right|< \delta,$$
$$\left|\frac{\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1}}{\lambda (pz^{p-1}-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1})+(\alpha-\gamma)}\right|< \delta\,.$$
Since \(\Re z\le\left|z\right|,\) then
$$\Re \left\lbrace\frac{\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1}}{\lambda (pz^{p-1}-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_kz^{k-1})+(\alpha-\gamma)}\right\rbrace < \delta\,. $$
If we choose \(z\) real and let \(z\rightarrow 1^{-}\), then we get
\begin{align*}& \sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k\le \delta\left[ \lambda (p-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k)+(\alpha-\gamma)\right],\\
& \Rightarrow \sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k \le \delta \lambda p -\delta \lambda\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k+\delta(\alpha-\gamma),\\
& \Rightarrow \sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k+\delta \lambda\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_k\le\delta(\lambda p+\alpha-\gamma),\\
& \Rightarrow \sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta \lambda)a_k \le\delta(\lambda p+\alpha-\gamma).\end{align*}
Conversely, suppose that the inequality (\ref{eq1.15}) holds true and that \(z\in \partial U:\left\lbrace z\in C:\left| z\right|=1\right\rbrace\)
and suppose that
\begin{align*}& \left|(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}\right| – \delta\left| \lambda (D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma)\right|\\
& \le\left|(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}- \delta(\lambda(D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma))\right|\\
& =\left|-\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_{k}z^{k-1} -\delta\lambda pz^{p-1} \right.\\&+\left. \delta\lambda\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_{k}z^{k-1}-\delta(\alpha-\gamma)\right|\\
& =\left|\sum_{k=p+1}^{\infty}(\delta\lambda-1)k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}a_{k}z^{k-1}-\delta(\lambda pz^{p-1}+\alpha-\gamma)\right|\\
& \le\sum_{k=p+1}^{\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}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(\delta\lambda+1)a_{k}-\delta(\lambda p+\alpha-\gamma)\le 0.\end{align*}
Since by maximum modulus theorem ,that the maximum modulus of an analytic function cannot be attained inside the domain but on the boundary, implies
$$\left|(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}\right|- \delta\left| \lambda (D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma)\right|< 0\,,$$
i.e.,$$ \left|(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}\right|< \delta\left| \lambda (D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma)\right|\,.$$
So, $$\frac{\left|(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}\right|}{\left|\lambda (D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma)\right|}< \delta\,,$$
implies
$$ \left|\frac{(D^{n}(\mu,\beta,t,p)f(z))'-pz^{p-1}}{\lambda (D^{n}(\mu,\beta,t,p)f(z))'+(\alpha-\gamma)}\right|< \delta.$$
Hence, we have that \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta).\)
Corollary 1.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), 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 \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) is a class of convex functions.
Proof
Let the functions
\begin{align}\tag{16}\label{eq1.16}
f(z)=z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k} ,&& (a_k\ge 0,p=1,2,…),
\end{align}
\begin{align}\tag{17}\label{eq1.17}
g(z)=z^{p}-\sum_{k=p+1}^{\infty}b_{k}z^{k} ,&& (b_k\ge 0,p=1,2,…),
\end{align}
be in the class \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then for \(0\le j\le1\)
$$h(z)=(1-j)f(z)+jg(z)=z^{p}-\sum_{k=p+1}^{\infty}c_{k}z^{k}\,,$$
where \(c_{k}=(1-j) a_{k}+j b_{k} \ge0\), then making use of (\ref{eq1.15}), we see that
\begin{align*}\sum_{k=p+1}^{\infty}k&[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)c_k\\
&=\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k+\sum_{k=p+1}^{\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{align*}
implies \(h(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), which completes the proof.
Theorem 3.
If each of the functions \(f(z)\) and \(g(z)\) is in the class \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then
\((f\ast g)(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\Omega)\),
where
\(\Omega \ge \frac{\delta^{2}(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)^{2}-\delta^{2}\lambda(\lambda p+\alpha-\gamma)}\).
Proof
From (\ref{eq1.15}), we have
\begin{align}\tag{18}\label{eq1.18}
\sum_{k=p+1}^{\infty}\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}a_{k}\le 1\,,
\end{align}
and
\begin{align}\tag{19}\label{eq1.19}
\sum_{k=p+1}^{\infty}\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}b_{k}\le 1\,.
\end{align}
We need to find the smallest \(\Omega\) such that
\begin{align}\tag{20}\label{eq1.20}
\sum_{k=p+1}^{\infty}\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\Omega\lambda)}{\Omega(\lambda p+\alpha-\gamma)}a_{k}b_{k}\le 1\,.
\end{align}
From (\ref{eq1.18}) and (\ref{eq1.19}), we find by means of Cauchy-Schwarz inequalities that
\begin{align}\tag{21}\label{eq1.21}
\sum_{k=p+1}^{\infty}\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\sqrt {a_{k}b_{k}}\le 1\,.
\end{align}
Thus, it is enough to show that
$$\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\Omega\lambda)}{\Omega(\lambda p+\alpha-\gamma)}a_{k}b_{k}\le \frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\sqrt {a_{k}b_{k}}\,.$$
That is
\begin{align}\tag{22}\label{eq1.22}
\sqrt{a_{k}b_{k}} \le \frac{\Omega(1+\delta \lambda)}{\delta(1+\Omega \lambda)}\,.
\end{align}
On the other hand, from (\ref{eq1.21}), we have
\begin{align}\tag{23}\label{eq1.23}
\sqrt{a_{k}b_{k}} \le \frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}\,.
\end{align}
Therefore, in view of (\ref{eq1.22}) and (\ref{eq1.23}), it is enough to show that
$$\frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}\le \frac{\Omega(1+\delta \lambda)}{\delta(1+\Omega \lambda)},$$
i.e., $$ \delta(\lambda p+\alpha-\gamma)\delta(1+\Omega \lambda)\le kM(1+\delta\lambda)\Omega(1+\delta \lambda)\,,$$
where \(M=[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}\). So,
$$\delta^{2}[(\lambda p+\alpha-\gamma)+\Omega \lambda(\lambda p+\alpha-\gamma)]\le k\Omega M(1+\delta \lambda)^{2}\,,$$
implies
$$ \delta^{2}(\lambda p+\alpha-\gamma)+\delta^{2}\Omega \lambda(\lambda p+\alpha-\gamma) \le k\Omega M(1+\delta \lambda)^{2}\,,$$
implies
\[ \delta^{2}(\lambda p+\alpha-\gamma) \le k\Omega M(1+\delta \lambda)^{2}-\delta^{2}\Omega \lambda(\lambda p+\alpha-\gamma)\,.\]
Also
\[\Omega\left[ k M(1+\delta \lambda)^{2}-\delta^{2} \lambda(\lambda p+\alpha-\gamma)\right] \ge \delta^{2}(\lambda p+\alpha-\gamma)\,,\]
implies
\[ \Omega \ge\frac{\delta^{2}(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)^{2}-\delta^{2}\lambda(\lambda p+\alpha-\gamma)}\,.\]
Theorem 4.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), 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{align*}r_{1}(\lambda,\alpha,\gamma,\delta,\rho)=\inf_{k}\left\lbrace\frac{[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)(p-\rho)}{\delta(\lambda p+\alpha-\gamma)} \right\rbrace^{\frac{1}{k-p}},& & (k\ge p+1)\,.\end{align*}
Proof
Let \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then
\[\left|\frac{f'(z)}{z^{p-1}}-p\right|< p-\rho\,,\]
implies
\begin{align}\tag{24}\label{eq1.24}
\left|\frac{pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1}-pz^{p-1}}{z^{p-1}}\right|=\left|\sum_{k=p+1}^{\infty}ka_{k}z^{k-p}\right|\le\sum_{k=p+1}^{\infty}ka_{k}\left|z\right|^{k-p}< p-\rho\,.
\end{align}
Since
\begin{align}\tag{25}\label{eq1.25}
\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k \le \delta(\lambda p+\alpha-\gamma)\,,
\end{align}
hence, (\ref{eq1.24}) is true if
\begin{align}\tag{26}\label{eq1.26}
\frac{k\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)}\,.
\end{align}
Solving (\ref{eq1.26}) for \(\left|z\right|\), we obtain
\begin{align*}
\left|z\right|< \left\lbrace\frac{[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)(p-\rho)}{\delta(\lambda p+\alpha-\gamma)} \right\rbrace^{\frac{1}{k-p}}, & & (k\ge p+1).\end{align*}
Hence, the proof.
Theorem 5.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\),then \(f(z)\) is p-valently starlike of order \(\rho\) in \(\left|z\right|< r_{2}(\lambda,\alpha,\gamma,\delta,\rho)\), where
\begin{align*}
r_{2}(\lambda,\alpha,\gamma,\delta,\rho)=\inf_{k}\left\lbrace\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)(p-\rho)}{\delta(\lambda p+\alpha-\gamma)(k-p)} \right\rbrace^{\frac{1}{k-p}},& & (k\ge p+1).\end{align*}
Proof
Let \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then
\[\left|\frac{zf'(z)}{f(z)}-p\right|< p-\rho\,.\]
The inequality
\begin{align}\tag{27}
\left|\frac{zf'(z)}{f(z)}-p\right|& =\left|\frac{z(pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1})-p(z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k})}{z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k}}\right|\notag\\
\label{eq1.27}& =\left|\frac{pz^{p}-\sum_{k=p+1}^{\infty}ka_{k}z^{k}-pz^{p}+p\sum_{k=p+1}^{\infty}a_{k}z^{k})}{z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k}}\right|\,.
\end{align}
Since
\begin{align}\tag{28}\label{eq1.28}
\left|\frac{-\sum_{k=p+1}^{\infty}(k-p)ka_{k}z^{k}}{z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k}}\right|=
\left|\frac{\sum_{k=p+1}^{\infty}(k-p)a_{k}z^{k-p}}{1-\sum_{k=p+1}^{\infty}a_{k}z^{k-p}}\right|\le \frac{\sum_{k=p+1}^{\infty}(k-p)a_{k}\left| z\right| ^{k-p}}{1-\sum_{k=p+1}^{\infty}a_{k}\left|z\right|^{k-p}}< p-\rho\,,
\end{align}
i.e.,
\begin{align*}
\sum_{k=p+1}^{\infty}ka_{k}\left| z\right| ^{k-p}-\sum_{k=p+1}^{\infty}pa_{k}\left| z\right| ^{k-p}&< (p-\rho)(1-\sum_{k=p+1}^{\infty}a_{k}\left|z\right|^{k-p})\\
& = \sum_{k=p+1}^{\infty}ka_{k}\left| z\right| ^{k-p}-\sum_{k=p+1}^{\infty}pa_{k}\left| z\right| ^{k-p}\end{align*}
\begin{align*}
& < p-\sum_{k=p+1}^{\infty}pa_{k}\left|z\right|^{k-p}-\rho+\sum_{k=p+1}^{\infty}\rho a_{k}\left|z\right|^{k-p}\\
& = \sum_{k=p+1}^{\infty}ka_{k}\left|z\right|^{k-p}-\sum_{k=p+1}^{\infty}\rho a_{k}\left|z\right|^{k-p}\\
&< p-\rho\\
& = \sum_{k=p+1}^{\infty}\frac {(k-\rho)a_{k}\left|z\right|^{k-p}}{p-\rho}\\& < 1.\end{align*}
Since
\[
\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k \le \delta(\lambda p+\alpha-\gamma)\,.
\]
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{align*}
\left|z\right|< \left\lbrace\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)(p-\rho)}{(k-p)\delta(\lambda p+\alpha-\gamma)} \right\rbrace^{\frac{1}{k-p}} , & & (k\ge p+1),\end{align*}
hence, the proof.
Theorem 6.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then \(f(z)\) is p-valently convex of order \(\rho\) in \(\left|z\right|< r_{3}(\lambda,\alpha,\gamma,\delta,\rho)\), where
\begin{align*}
r_{3}(\lambda,\alpha,\gamma,\delta,\rho)=\inf_{k}\left\lbrace\frac{[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)p(p-\rho)}{\delta(\lambda p+\alpha-\gamma)(k-p)} \right\rbrace^{\frac{1}{k-p}}, & & (k\ge p+1).\end{align*}
Proof
Let \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then
\[\left|1+\frac{zf”(z)}{f'(z)}-p\right|< p-\rho\,.\]
The inequality
\begin{align*}
& \left|1+\frac{zf''(z)}{f'(z)}-p\right|\\
& =\left|\frac{pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1}+z(p(p-1)z^{p-2}-\sum_{k=p+1}^{\infty}k(k-1)a_{k}z^{k-2})-p(pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1})}{pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1}}\right|\\
& =\left|\frac{pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1}+p(p-1)z^{p-1}-\sum_{k=p+1}^{\infty}k(k-1)a_{k}z^{k-1})-p^{2}z^{p-1}+\sum_{k=p+1}^{\infty}pka_{k}z^{k-1})}{pz^{p-1}-\sum_{k=p+1}^{\infty}ka_{k}z^{k-1}}\right|\\
& =\left|\frac{-\sum_{k=p+1}^{\infty}k(k-p)a_{k}z ^{k-p}}{p-\sum_{k=p+1}^{\infty}ka_{k}z^{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*}
So,
\[\sum_{k=p+1}^{\infty}k(k-p)a_{k}\left| z\right| ^{k-p}< (p-\rho)(p-\sum_{k=p+1}^{\infty}ka_{k}\left|z\right|^{k-p})\,,\]
implies
\[\sum_{k=p+1}^{\infty}k(k-p)a_{k}\left| z\right| ^{k-p}< p(p-\rho)-(p-\rho)\sum_{k=p+1}^{\infty}ka_{k}\left|z\right|^{k-p})\,,\]
implies
\[\sum_{k=p+1}^{\infty}k(k-\rho)a_{k}\left| z\right|^{k-p}< p(p-\rho)\,.\]
Since
\(
\sum\limits_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k \le \delta(\lambda p+\alpha-\gamma)
\).
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{align*}
\left|z\right|< \left\lbrace\frac{[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)p(p-\rho)}{(k-p)\delta(\lambda p+\alpha-\gamma)} \right\rbrace^{\frac{1}{k-p}},& & (k\ge p+1),\end{align*}
hence, the proof.
Theorem 7.
Let
\begin{align}\tag{29}\label{eq1.29}
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{align}
then, \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) if and only if it can be expressed in the form
\[f(z)=\lambda_{p}f_{p}(z)+\sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z)\,,\]
where \(\lambda_{p}\ge 0\) and \(\lambda_{p}\)=\(1 – \sum\limits_{k=p+1}^{\infty}\lambda_{k}\).
Proof
Assume that
\(f(z)=\lambda_{p}f_{p}(z)+\sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z)\), then
\begin{align}\tag{30}\label{eq1.30}
f(z)=(1 – \sum_{k=p+1}^{\infty}\lambda_{k})z^{p}+\sum_{k=p+1}^{\infty}\lambda_{k}\left\{ z^{p}-\frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}z^{k}\right\}\,,
\end{align}
implies
\[f(z)=z^{p}-\sum_{k=p+1}^{\infty}\lambda_{k}\left\{
\frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}\right\}z^{k}\,.\]
Thus,
\begin{align*}
&\sum_{k=p+1}^{\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}^{\infty}\lambda_{k}
= \delta(\lambda p+\alpha-\gamma)(1-\lambda_{p}) \le \delta(\lambda p+\alpha-\gamma)\,,\end{align*}
which shows that \(f(z)\) satisfies condition (\ref{eq1.15}) and therefore, \(f\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta).\)
Conversely, suppose that \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), since
\begin{align*} 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{align*}
we may set
\begin{align*}
& \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}^{\infty}\lambda_{k}\,,\end{align*}
then we obtain from \(f(z)= z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k}\),
\[ f(z)=(\lambda_{p}+\sum_{k=p+1}^{\infty}\lambda_{k})z^{p}- \sum_{k=p+1}^{\infty}\lambda_{k} \frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}z^{k}\,,\]
i.e.,
\[ f(z)=\lambda_{p}z^{p}+ \sum_{k=p+1}^{\infty}\lambda_{k}(z^{p}- \frac{\delta(\lambda p+\alpha-\gamma)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}z^{k})\,,\]
implies
\[ f(z)=\lambda_{p}z^{p} + \sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z)\,.\]
This completes the proof.
Corollary 2.
The extreme points of \(S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) are given by;
\begin{align*}
& 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{align*}
Theorem 8.
Let \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) and suppose that
\begin{align}\tag{31}\label{eq1.31}
\sum_{j=p+1}^{\infty}(j-q)_{q+1}a_{j} \le \frac{\delta(\lambda p+\alpha-\gamma) \Gamma (k+1)\Gamma (2+p-l-q)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (k+1-l-q)\Gamma (p+1-q)}\,,
\end{align}
for some \(0\le q \le j\), \(0\le l< 1\), \((j-q)_{q+1}\) denotes the pochhammer symbol defined by \((j-q)_{q+1}=(j-q)(j-q+1)…j.\) Also, let the function
\begin{align}\tag{32}\label{eq1.32}
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{align}
If there exists an analytic function \(w(z)\) defined by
\begin{align}\tag{33}\label{eq1.33}
(w(z))^{k-p}=\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma (k+1-l-q)}{\Gamma (k+1)}\sum_{j=p+1}^{\infty}(j-q)_{q+1\psi (j) a_{j}z^{j-p}}\,,
\end{align}
with \((k\ge q)\)
\begin{align*}\Psi (j) =\frac{\Gamma (j-q)}{\Gamma (j+1-l-q)},& & (0\le l0\) and \(z=re^{i\theta}\), \((0 < r< 1)\),
\begin{align}\tag{34}\label{eq1.34}
\int_{0}^{2\pi}\left|D_{z}^{q+l}f(z)\right|^{\sigma}d\theta \le \int_{0}^{2\pi}\left|D_{z}^{q+l}f_{k}(z)\right|^{\sigma}d\theta\,.
\end{align}
Proof
Let \(f(z)=z^{p}-\sum_{j=p+1}^{\infty}a_{j}z^{j}.\) By means of (\ref{eq1.12}) and Definition 6, we have
\begin{align}
D_{z}^{q+l}f(z)& =\frac{\Gamma (p+1)z^{p-l-q}}{\Gamma (p+1-l-q)}-\sum_{j=p+1}^{\infty}\frac{\Gamma (j+1)}{\Gamma (j+1-l-q)}a_{j}z^{j-l-q}\notag\\
&= \frac{\Gamma (p+1)z^{p-l-q}}{\Gamma (p+1-l-q)}[1-\sum_{j=p+1}^{\infty}\frac{\Gamma (j+1)\Gamma (p+1-l-q)}{\Gamma (p+1)\Gamma (j+1-l-q)}a_{j}z^{j-p}]\tag{35}\\
\label{eq1.35}
&= \frac{\Gamma (p+1)z^{p-l-q}}{\Gamma (p+1-l-q)}[1-\sum_{j=p+1}^{\infty}\frac{\Gamma (p+1-l-q)}{\Gamma (p+1) }(j-q)_{q+1}\Psi(j)a_{j}z^{j-p}\tag{36}]\,,
\end{align}
where
\( \Psi(j)=\frac{\Gamma (j-q)}{\Gamma (j+1-l-q)},\;\;\;\;(0\le l< 1, j\ge p+1).\)
Since \(\psi\) is a decreasing function of \(j\), we get
\[ 0< \Psi(j) \le \Psi(p+1)=\frac{\Gamma (p+1-q)}{\Gamma (2+p-l-q)}\,.\]
Similarly, from (\ref{eq1.32}), (\ref{eq1.14}), and Definition 6, we have
\begin{align}
D_{z}^{q+l}f_{k}(z)&=\frac{\Gamma (p+1)z^{p-l-q}}{\Gamma (p+1-l-q)}-\frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (k+1-l-q)}z^{k-l-q}\notag\\
\label{eq1.36}
&=\frac{\Gamma (p+1)z^{p-l-q}}{\Gamma (p+1-l-q)}[1-\frac{\delta(\lambda p+\alpha-\gamma)\Gamma(k+1)\Gamma (p+1-l-q)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (p+1)\Gamma (k+1-l-q)}z^{k-p}]\,\tag{37}.
\end{align}
For some \(\sigma>0\) and \(z=re^{i\theta}\), \((0 < r< 1)\), we show that
\begin{align}\label{eq1.37}
\int_{0}^{2\pi}&\left|1-\sum_{j=p+1}^{\infty}\frac{\Gamma (p+1-l-q)}{\Gamma (p+1)}(j-q)_{q+1}\psi(j)a_{j}z^{j-p}\right|^{\sigma}d(\theta)\notag\\
&\le \int_{0}^{2\pi}\left|1-\frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)\Gamma (p+1-l-q)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (p+1)\Gamma (k+1-l-q)}z^{k-p}\right|^{\sigma}d(\theta)\,,\tag{38}
\end{align}
so, by applying Lemma 1, it is enough to show that
\begin{align}\label{eq1.38}
1-&\sum_{j=p+1}^{\infty}\frac{\Gamma (p+1-l-q)}{\Gamma
(p+1)}(j-q)_{q+1}\psi(j)a_{j}z^{j-p} \notag\\
&\prec 1-\frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)\Gamma (p+1-l-q)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (p+1)\Gamma (k+1-l-q)}z^{k-p}\,\tag{39}.
\end{align}
If the above subordination holds true, then we have an analytic function \(w(z)\)
with \(w(0)=0\), \(|w(z)|< 1\), such that
\begin{align}\label{eq1.39}
1-&\sum_{j=p+1}^{\infty}\frac{\Gamma (p+1-l-q)}{\Gamma (p+1)}(j-q)_{q+1}\psi(j)a_{j}z^{j-p} \notag\\
&= 1-\frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)\Gamma (p+1-l-q)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (p+1)\Gamma (k+1-l-q)}w(z)^{k-p}\,\tag{40}.
\end{align}
By the condition of the Theorem, we define the function \(w(z)\) by
\begin{align}\tag{41}\label{eq1.40}
(w(z))^{k-p}=\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma(k+1-l-q)}{\Gamma (k+1)}\sum_{j=p+1}^{\infty}(j-q)_{q+1}\psi(j)a_{j}z^{j-p}\,,
\end{align}
which readily yields \(w(0)=0\). For such a function \(w(z)\), we have
\begin{align}\tag{42}
\notag\left|(w(z))\right|^{k-p}&\le \frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma(k+1-l-q)}{\Gamma (k+1)}\sum_{j=p+1}^{\infty}(j-q)_{q+1}\psi(j)a_{j}\left| z\right| ^{j-p} \\
\notag&\le \left| z\right| \frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma(k+1-l-q)}{\Gamma (k+1)}\psi(p+1)\sum_{j=p+1}^{\infty}(j-q)_{q+1}a_{j}\\
\notag&= \left| z\right| \frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma(k+1-l-q)\Gamma(p+1-q)}{\Gamma (k+1)\Gamma(2+p-l-q)}\sum_{j=p+1}^{\infty}(j-q)_{q+1}a_{j}\\
\label{eq1.41}
&\le \left|z\right|< 1.
\end{align}
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 \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) and suppose that
\begin{align}\tag{43}\label{eq1.42}
\sum_{j=p+1}^{\infty}ja_{j} \le \frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)\Gamma (2+p-l)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)
\Gamma (k+1-l)\Gamma (p+1)},& & (j\ge p+1)\,,
\end{align}
if there exists an analytic function \(w(z)\) defined by
\begin{align}\tag{44}\label{eq1.43}
(w(z))^{k-p}=\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma (k+1-l)}{\Gamma (k+1)}\sum_{j=p+1}^{\infty}{j} \Psi (j) a_{j}z^{j-p}\,,
\end{align}
with
\begin{align*}\psi(j)=\frac{\Gamma(j)}{\Gamma(j+1-l)},& & (0\le l0\) and \(z=re^{i\theta}\),\;\;\; \((0 < r < 1)\)
\begin{align}\tag{45}\label{eq1.44}
\int_{0}^{2\pi}\left|D_{z}^{l}f(z)\right|^{\sigma}d\theta \le \int_{0}^{2\pi}\left|D_{z}^{l}f_{k}(z)\right|^{\sigma}d\theta\,.
\end{align}
Letting \(q=1\), we have the following from Theorem 8.
Corollary 4.
Let \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\) and suppose that
\begin{align}\tag{46}\label{eq1.45}
\sum_{j=p+1}^{\infty}j(j-1)a_{j} \le \frac{\delta(\lambda p+\alpha-\gamma)\Gamma (k+1)\Gamma (p+1-l)}{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)\Gamma (k-l)\Gamma (p)}, & &(j\ge p+1)\,,
\end{align}
if there exists an analytic function \(w(z)\) define by
\begin{align}\tag{47}\label{eq1.46}
(w(z))^{k-p}=\frac{k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)}{\delta(\lambda p+\alpha-\gamma)}\frac{\Gamma (k-l)}{\Gamma (k+1)}\sum_{j=p+1}^{\infty}j(j-1) \Psi(j) a_{j}z^{j-p}
\,,\end{align}
with
\begin{align*} \psi(j)=\frac{\Gamma(j-1)}{\Gamma(j-l)},& & (0\le l0\) and \(z=re^{i\theta}\), \((0 < r < 1)\)
\begin{align}\tag{48}\label{eq1.47}
\int_{0}^{2\pi}\left|D_{z}^{1+l}f(z)\right|^{\sigma}d\theta \le \int_{0}^{2\pi}\left|D_{z}^{1+l}f_{k}(z)\right|^{\sigma}d\theta,& & (0\le l < 1).
\end{align}
Theorem 9.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then we have
\[\left|D_{z}^{-l}f(z)\right|\le\frac{\Gamma(p+1)}{\Gamma(p+l+1)}\left|z\right|^{p+l}\left[1+\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)(p+l+1)}\left|z\right|\right]\,,\]
and
\begin{align}\tag{49}\label{eq1.48}
\left|D_{z}^{-l}f(z)\right|\ge\frac{\Gamma(p+1)}{\Gamma(p+l+1)}\left|z\right|^{p+l}\left[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|\right]\,.
\end{align}
Proof
Suppose that \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), using Theorem 1, we find that
\[
\sum_{k=p+1}^{\infty}k[1+(\frac{k}{p}+\beta-\mu-1)t]^{n}(1+\delta\lambda)a_k \le \delta(\lambda p+\alpha-\gamma)\,,
\]
implies
\begin{align}\tag{50}\label{eq1.49}
\frac{(p+1)[p+(p(\beta-\mu)+1)t]^{n}(1+\delta\lambda)}{p^{n}}\sum_{k=p+1}^{\infty}a_k \le \delta(\lambda p+\alpha-\gamma)\,,
\end{align}
i.e.,
\begin{align}\tag{51}\label{eq1.50}
\sum_{k=p+1}^{\infty}a_k \le \frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{(p+1)[p+(p(\beta-\mu)+1)t]^{n}(1+\delta\lambda)}\,.
\end{align}
From (\ref{eq1.7}), if \(f(z)=z^{p}-\sum\limits_{k=p+1}^{\infty}a_{k}z^{k}\),
\[D_{z}^{-l}f(z)=\frac{\Gamma(p+1)}{\Gamma(p+l+1)}z^{p+l}-\sum_{k=p+1}^{\infty}\frac{\Gamma(k+1)}{\Gamma(k+l+1)}a_{k}z^{k+l}\,,\]
implies
\begin{align}\tag{52}\label{eq1.51}
\frac {\Gamma(p+l+1)}{\Gamma(p+1)}z^{-l}D_{z}^{-l}f(z)=z^{p}-\sum_{k=p+1}^{\infty}\frac{\Gamma(k+1)\Gamma(p+l+1)}{\Gamma(p+1)\Gamma(k+l+1)}a_{k}z^{k}= z^{p}-\sum_{k=p+1}^{\infty}\Psi(k)a_{k}z^{k}\,,
\end{align}
where
\begin{align}\tag{53}\label{eq1.52}
\Psi(k)=\frac{\Gamma(k+1)\Gamma(p+l+1)}{\Gamma(p+1)\Gamma(k+l+1)}\,.
\end{align}
Clearly, \(\psi\) is a decreasing function of \(k\) and we get
\[0< \Psi(k)\le\Psi(p+1)=\frac{p+1}{p+l+1}\,.\]
Using (\ref{eq1.50}) and (\ref{eq1.52}), we obtain,
\begin{align*}
\left|\frac{\Gamma(p+l+1)}{\Gamma(p+1)}z^{-l}D_{z}^{-l}f(z)\right|&\le\left|z\right|^{p}+\psi(p+1)\left|z\right|^{p+1}\sum_{k=p+1}^{\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{align*}
which is equivalent to assertion (\ref{eq1.48}) and
\begin{align*}
\left|\frac{\Gamma(p+l+1)}{\Gamma(p+1)}z^{-l}D_{z}^{-l}f(z)\right|&\ge\left|z\right|^{p}-\psi(p+1)\left|z\right|^{p+1}\sum_{k=p+1}^{\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{align*}
which completes the proof.
Theorem 10.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then we have
\[\left|D_{z}^{l}f(z)\right|\le\frac{\Gamma(p+1)}{\Gamma(p-l+1)}\left|z\right|^{p-l}\left[1+\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)(p-l+1)}\left|z\right|\right]\,,\]
and
\begin{align}\tag{54}\label{eq1.53}
\left|D_{z}^{l}f(z)\right|\ge\frac{\Gamma(p+1)}{\Gamma(p-l+1)}\left|z\right|^{p-l}\left[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|\right]\,.
\end{align}
Proof
If \(f(z)=z^{p}-\sum\limits_{k=p+1}^{\infty}a_{k}z^{k}\), then
\[D_{z}^{l}f(z)=\frac{\Gamma(p+1)}{\Gamma(p-l+1)}z^{p-l}-\sum_{k=p+1}^{\infty}\frac{\Gamma(k+1)}{\Gamma(k-l+1)}a_{k}z^{k-l}\,,\]
implies
\begin{align}\tag{55}\label{eq1.54}
\frac {\Gamma(p-l+1)}{\Gamma(p+1)}z^{l}D_{z}^{l}f(z)=z^{p}-\sum_{k=p+1}^{\infty}\frac{\Gamma(k+1)\Gamma(p-l+1)}{\Gamma(p+1)\Gamma(k-l+1)}a_{k}z^{k}= z^{p}-\sum_{k=p+1}^{\infty}\Psi(k)a_{k}z^{k}\,,
\end{align}
where
\begin{align}\tag{56}\label{eq1.55}
\Psi(k)=\frac{\Gamma(p-l+1)\Gamma(k+1)}{\Gamma(p+1)\Gamma(k-l+1)}\,.
\end{align}
Clearly, \(\Psi\) is a decreasing function of \(k\) and we get
\[0< \Psi(k)\le\Psi(p+1)=\frac{p+1}{p-l+1}\,.\]
Using (\ref{eq1.50}) and (\ref{eq1.55}), we obtain,
\begin{align*}
\left|\frac{\Gamma(p-l+1)}{\Gamma(p+1)}z^{l}D_{z}^{l}f(z)\right|&\le\left|z\right|^{p}+\psi(p+1)\left|z\right|^{p+1}\sum_{k=p+1}^{\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{align*}
which is equivalent to assertion (\ref{eq1.53}).
Similarly, \begin{align*}
\left|\frac{\Gamma(p-l+1)}{\Gamma(p+1)}z^{l}D_{z}^{l}f(z)\right|&\ge\left|z\right|^{p}-\Psi(p+1)\left|z\right|^{p+1}\sum_{k=p+1}^{\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{align*}
which completes the proof.
Corollary 5.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then we have
\begin{align}\tag{57}\label{eq1.56}
\notag \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\left|f(z)\right|\\
&\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{align}
Proof
From Definition 4, we have
\[\lim_{l\to 0} D_{z}^{-l}f(z)=f(z)\,.\]
Therefore, setting \(l=0\) in (\ref{eq1.48}), we obtain
\[\left|D_{z}^{0}f(z)\right|\le\frac{\Gamma(p+1)}{\Gamma(p+0+1)}\left|z\right|^{p+0}\left[1+\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)(p+0+1)}\left|z\right|\right]\,,\]
and
\begin{align*}%\label{eq1.57}
\left|D_{z}^{0}f(z)\right|\ge\frac{\Gamma(p+1)}{\Gamma(p+0+1)}\left|z\right|^{p+0}\left[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|\right]\,,
\end{align*}
i.e.,
\begin{align*}%\label{eq1.58}
\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\left|f(z)\right|\notag\\
&\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{align*}
which is (57).
Corollary 6.
If \(f(z)\in S^{n}_{p}(\lambda,\alpha,\gamma,\delta)\), then we have
\begin{align}\tag{58}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|^{}\le\left|f'(z)\right|
\label{eq1.59}
\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}\,.
\end{align}
Proof
From Definition 4, we have
\[\lim_{l\to 1} D_{z}^{l}f(z)=f'(z)\,.\]
Therefore, setting \(l=1\) in (\ref{eq1.53}), we obtain
\[\left|D_{z}^{1}f(z)\right|\le\frac{\Gamma(p+1)}{\Gamma(p)}\left|z\right|^{p-1}\left[1+\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)(p)}\left|z\right|\right]\,,\]
and
\begin{align*}
\left|D_{z}^{1}f(z)\right|\ge\frac{\Gamma(p+1)}{\Gamma(p)}\left|z\right|^{p-1}\left[1-\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)(p)}\left|z\right|\right]\,,
\end{align*}
i.e., \[ p\left|z\right|^{p-1}-\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)}\left|z\right|^{p}\le\left|f'(z)\right|
\le p\left|z\right|^{p-1}+\frac{\delta(\lambda p+\alpha-\gamma)p^{n}}{[p+(p(\beta-\mu)+1)t]^{n}(1+\delta \lambda)}\left|z\right|^{p}\,,\]
which is (\ref{eq1.59}).
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.
