Let $\Sigma_{p}$ denote the class of function $f$ of the form:

$\begin{equation}\label{1.1} f(z)=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}a_{k}z^{k}, p\in \mathbb{N}=\{1,2,3,\cdots\}, \end{equation}$

(1.1)

which are analytic in the punctured open unit disk

$\mathbb{U}^{*}:=\{z:z\in \mathbb{C}\ {\rm and}\ 0<|z|<1 \}=:\mathbb{U}\setminus \{0\} . $

A function $f\in \Sigma_{p}$ is said to be in the class $ \mathcal{MS}_{p}^{*}(\alpha)$ of meromorphic $p$-valent starlike function of order $\alpha$ if it satisfies the inequality $\Re\left(\frac{zf'(z)}{f(z)}\right)<-\alpha(z\in\mathbb{U};\ 0\leqq\alpha<p). $

Let $\mathcal{P}$ denote the class of functions $p$ given by

$\begin{equation}\label{1.2} p(z)=1+\sum\limits^{\infty}_{k=1}p_{k}z^{k}(z\in \mathbb{U}), \end{equation}$

(1.2)

which are analytic in $\mathbb{U}$ and satisfy the condition $\Re\big(p(z)\big) >0(z\in \mathbb{U}).$

Given two functions $f,\ g\in\Sigma_{p}$, where $f$ is given by (1.1) and $g$ is given by

$g(z)=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}b_{k}z^{k},$

the Hadamard product $f*g$ is defined by

$(f*g)(z):=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}a_{k}b_{k}z^{k}=:(g*f)(z).$

A functions $f\in \Sigma_{p}$ is said to be in the class $\mathcal{H}_{p}(\beta,\lambda)$ if it satisfies the condition

$\begin{equation}\label{1.3} \Re\left(\frac{zf'(z)}{f(z)}+\beta\frac{z^2f''(z)}{f(z)}\right)< \beta\lambda \left(\lambda+\frac{1}{2}\right)+\frac{1}{2}p\beta-\lambda\left(z\in\mathbb{U}\right), \end{equation}$

(1.3)

where (and throughout this paper unless otherwise mentioned) the parameters $\beta$ and $\lambda$ are constrained as follows:

$\begin{equation}\label{1.4} \beta\geqq 0\qquad {\rm and} \qquad p-\frac{1}{2}\leqq \lambda <p. \end{equation}$

(1.4)

Clearly, we have $\mathcal{H}_{p}(0,\lambda)=\mathcal{MS}_{p}^{*}(\lambda).$

In order to prove our main results, we need the following lemmas.

Lemma 2.1 (see [2]) If the function $p\in \mathcal{P}$ is given by (1.2), then $|p_{k}|\leqq 2(k\in \mathbb{N}).$

The following lemma shows that the class $\mathcal{H}_{p}(\beta,\lambda)$ is a subclass of the class $\mathcal{MS}_{p}^{*}(\lambda)$ of meromorphic $p$-valent starlike functions of order $\lambda.$

Lemma 2.2 (see [1]) If $f\in \mathcal{H}_{p}(\beta,\lambda)$, then $f\in \mathcal{MS}_{p}^{*}(\lambda)$.

Lemma 2.3 Let $\beta>0$ and $p-\beta p(p+1)-\gamma>0$. Suppose also that the sequence $\{A_{k}\}^{\infty}_{k=1}$ is defined by

$\begin{equation}\label{2.1} A_{1}=\frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1} ,\end{equation}$

(2.1)

and

$\begin{equation}\label{2.1111}A_{k+1}=\frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+ (\beta k+1)(k+1)}\left(1+\sum\limits^{k}_{l=1}|A_{l}|\right)\qquad(k\in \mathbb{N}).\end{equation}$

(2.2)

Then

$\begin{equation}\label{2.2} A_{k}= \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1}\prod\limits^{k-1}_{j=1}\frac{3p-3\beta p(p+1)-2\gamma+(\beta j+1-\beta)j}{p-\beta p(p+1)+(\beta j+1)(j+1)} \qquad(k\in \mathbb{N}\setminus\{1\}). \end{equation}$

(2.3)

Proof By virtue of (2.1), we easily get

$\begin{equation}\label{2.3} [p-\beta p(p+1)+ (\beta k+1)(k+1)]A_{k+1}=2(p-\beta p(p+1)-\gamma)\left(1+\sum\limits^{k}_{l=1}|A_{l}|\right), \end{equation}$

(2.4)

and

$\begin{equation}\label{2.4} [p-\beta p(p+1)+ (\beta k+1-\beta)k]A_{k}=2(p-\beta p(p+1)-\gamma)\left(1+\sum\limits^{k-1}_{l=1}|A_{l}|\right). \end{equation}$

(2.5)

Combining (2.4) and (2.5), we obtain

$\begin{equation}\label{2.5} \frac{A_{k+1}}{A_{k}}= \frac{3p-3\beta p(p+1)-2\gamma+(\beta k+1-\beta)k}{p-\beta p(p+1)+(\beta k+1)(k+1)}. \end{equation}$

(2.6)

Thus, for $k\geqq 2$, we deduce from (2.6) that

$\begin{align*}\begin{split}A_{k}&=\frac{A_{k}}{A_{k-1}}\cdots \frac{A_{3}}{A_{2}}\cdot\frac{A_{2}}{A_{1}}\cdot A_{1} \\&=\frac{2(p-\beta p(p+1)-\gamma)}{p-\beta p(p+1)+1}\prod^{k-1}_{j=1}\frac{3p-3\beta p(p+1)-2\gamma+(\beta j+1-\beta)j}{p-\beta p(p+1)+(\beta j+1)(j+1)}.\end{split}\end{align*}$

Theorem 3.1 Let

$\begin{equation}\label{3.1} p-\beta p(p+1)+\beta\lambda\left(\lambda+\frac{1}{2}\right)+ \frac{1}{2}p\beta-\lambda>0. \end{equation}$

(3.1)

Suppose also that $f\in \Sigma_{p}$ is given by (1.1). If

$\begin{equation}\label{3.2} \sum\limits^{\infty}_{k=1}[k+\beta k(k-1)+\gamma]|a_{k}|\leqq p-\beta p(p+1)-\gamma, \end{equation}$

(3.2)

where (and throughout this paper unless otherwise mentioned) the parameter $\gamma$ is constrained as follows:

$\begin{equation}\label{3.3} \gamma:=\lambda-\beta\lambda\left(\lambda+\frac{1}{2}\right)-\frac{1}{2}p\beta, \end{equation}$

(3.3)

then $f\in \mathcal{H}_{p}(\beta,\lambda)$.

Proof To prove $f\in \mathcal{H}_{p}(\beta,\lambda)$, it suffices to show that

$\begin{equation}\label{3.4} \left| \frac{\frac{zf^{'}(z)}{f(z)}+\beta\frac{z^{2}f^{''}(z)}{f(z)} +1}{\frac{zf^{'}(z)}{f(z)}+\beta\frac{z^{2}f^{''}(z)}{f(z)} +2\gamma-1}\right|<1\qquad(z\in \mathbb{U}). \end{equation}$

(3.4)

Combining (3.1), (3.2) and (3.3), we know that

$\begin{align}\label{3.5}& p-\beta p(p+1)-2\gamma+1- \sum\limits^{\infty}_{k=1}[k+\beta k(k-1)+2\gamma-1]|a_{k}|\\\notag \geqq &1-p+\beta p(p+1)+ \sum\limits^{\infty}_{k=1}[k+1+\beta k(k-1)]|a_{k}|>0.\end{align}$

(3.5)

Now, by the maximum modulus principle, we deduce from (1.1) and (3.5) that

$\begin{align*} &\left| \frac{\frac{zf^{'}(z)}{f(z)}+\beta\frac{z^{2}f^{''}(z)}{f(z)} +1}{\frac{zf^{'}(z)}{f(z)}+\beta\frac{z^{2}f^{''}(z)}{f(z)}+ 2\gamma-1}\right| \\ =&\left| \frac{1-p+\beta p(p+1)+ \sum^{\infty}_{k=1}[k+1+\beta k(k-1)]a_{k}z^{k+p}}{ -p+\beta p(p+1)+2\gamma-1+ \sum^{\infty}_{k=1}[k+\beta k(k-1)+2\gamma-1]a_{k}z^{k+p}} \right|\\ <&\frac{1-p+\beta p(p+1)+ \sum^{\infty}_{k=1}[k+1+\beta k(k-1)]|a_{k}|}{ p-\beta p(p+1)-2\gamma+1- \sum^{\infty}_{k=1}[k+\beta k(k-1)+2\gamma-1]|a_{k}|}\\ \leqq &1. \end{align*}$

Theorem 3.2 Let $\gamma$ be defined by (3.3). If $f\in \mathcal{H}_{p}(\beta,\lambda)$ with $0<\beta<\frac{p-\lambda}{p(p+1)}$, then

$|a_{1}|\leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1},$

and

$|a_{k}|\leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1}\prod\limits^{k-1}_{j=1}\frac{3p-3\beta p(p+1)-2\gamma+(\beta j+1-\beta)j}{p-\beta p(p+1)+(\beta j+1)(j+1)} \qquad(k\in \mathbb{N}\setminus\{1\}).$

Proof Suppose that

$\begin{equation}\label{3.6} q(z):=-\frac{zf^{'}(z)}{f(z)}-\beta\frac{z^{2}f^{''}(z)}{f(z)}+\beta\lambda(\lambda+\frac{1}{2})+\frac{1}{2}p\beta-\lambda . \end{equation}$

(3.6)

Then, $q$ is analytic in $\mathbb{U}$ and $\Re\left(q(z)\right)>0(z\in\mathbb{U})$ with $q(0)=p-\beta p(p+1)-\gamma>0.$ It follows from (3.3) and (3.6) that

$\begin{equation}\label{3.7} q(z)f(z)=-zf^{'}(z)-\beta z^{2}f^{''}(z)-\gamma f(z). \end{equation}$

(3.7)

By noting that

$h(z)=\frac{q(z)}{p-\beta p(p+1)-\gamma}\in \mathcal{P},$

if we put

$q(z)=c_{0}+\sum\limits^{\infty}_{k=1}c_{k}z^{k}\qquad (c_{0}=p-\beta p(p+1)-\gamma).$

by Lemma 2.1, we know that $|c_{k}|\leqq 2[p-\beta p(p+1)-\gamma](k\in \mathbb{N}).$ It follows from (3.7) that

$\begin{align}\begin{split}\label{3.8} &\left(c_{0}+\sum\limits^{\infty}_{k=1}c_{k}z^{k}\right)\left(\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}a_{k}z^{k}\right) \\=&\left(\frac{p}{z^{p}}-\sum\limits^{\infty}_{k=1}ka_{k}z^{k}\right)-\left[\beta\frac{p(p+1)}{z^{p}}+\beta\sum\limits^{\infty}_{k=1}k(k-1)a_{k}z^{k}\right] -\gamma\left(\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}a_{k}z^{k}\right). \end{split}\end{align}$

(3.8)

In view of (3.8), we get

$\begin{equation}\label{3.9} c_{0}a_{1}+c_{p+1}=-a_{1}-\gamma a_{1}, \end{equation}$

(3.9)

and

$\begin{equation}\label{3.10} c_{0}a_{k+1}+c_{p+k+1}+\sum\limits^{k}_{l=1}a_{l}c_{k+1-l}=-(k+1)a_{k+1}-\beta k(k+1)a_{k+1}-\gamma a_{k+1}. \end{equation}$

(3.10)

From (3.9), we obtain

$\begin{equation}\label{3.11} |a_{1}|\leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1}, \end{equation}$

(3.11)

and

$\begin{equation}\label{3.12} |a_{k+1}|\leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+ (\beta k+1)(k+1)}\left(1+\sum\limits^{k}_{l=1}|A_{l}|\right)\qquad(k\in \mathbb{N}). \end{equation}$

(3.12)

Next, we define the sequence $\{A_{k}\}^{\infty}_{k=1}$ as follows

$\begin{equation}\label{3.13} A_{1}=\frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1},\end{equation}$

(3.13)

and

$\begin{equation}A_{k+1}=\frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+ (\beta k+1)(k+1)}\left(1+\sum\limits^{k}_{l=1}|A_{l}|\right)\qquad(k\in \mathbb{N}).\end{equation}$

(3.14)

In order to prove that

$ |a_{k}|\leqq |A_{k}|\qquad(k\in \mathbb{N}) ,$

we make use of the principle of mathematical induction. By noting that

$|a_{1}|\leqq | A_{1}|=\frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+1}.$

Therefore, assuming that

$|a_{l}|\leqq | A_{l}|\qquad(l=1,2,3,\cdots,k;\ k\in \mathbb{N}).$

Combining (3.12) and (3.13), we get

$\begin{align*} |a_{k+1}|&\leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+ (\beta k+1)(k+1)}\left(1+\sum\limits^{k}_{l=1}|a_{l}|\right)\\ & \leqq \frac{2[p-\beta p(p+1)-\gamma]}{p-\beta p(p+1)+ (\beta k+1)(k+1)}\left(1+\sum\limits^{k}_{l=1}|A_{l}|\right)\\ &=A_{k+1}\qquad(k\in \mathbb{N}). \end{align*}$

Hence, by the principle of mathematical induction, we have

$\begin{equation}\label{3.14} |a_{k}|\leqq | A_{k}|\qquad(k\in \mathbb{N}) \end{equation}$

(3.15)

as desired.

By means of Lemma 2.3 and (3.13), we know that (2.3) holds true. Combining (3.15) and (2.3), we readily get the coefficient estimates asserted by Theorem 3.2. Assuming that $\gamma$ is given by (3.3) and that the condition (3.1) holds true, we here introduce the $\delta$-neighborhood of a function $ f\in\Sigma_{p}$ of the form (1.1) by means of the following definition:

$\begin{align}\label{3.15} \mathcal{N}_{\delta}(f):=\left\{g\in\Sigma_{p}: g(z)=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}b_{k}z^{k}\ {\rm and} \ \sum\limits^{\infty}_{k=1}\frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}|a_{k}-b_{k}|\leqq\delta \right\}, \end{align}$

(3.16)

where $\delta\geqq0$.

Theorem 3.3 Let the condition (3.1) hold true. If $ f\in\Sigma_{p}$ satisfies the condition

$\begin{equation}\label{3.16} \frac{f(z)+\varepsilon z^{-p}}{1+\varepsilon} \in \mathcal{H}_{p}(\beta,\lambda)\qquad(\varepsilon \in \mathbb{C};\ |\varepsilon|<\delta;\ \delta>0), \end{equation}$

(3.17)

then

$\begin{equation}\label{3.17} \mathcal{N}_{\delta}(f)\subset \mathcal{H}_{p}(\beta,\lambda). \end{equation}$

(3.18)

Proof By noting that the condition (1.3) is equivalent to (3.4), we easily find from (3.4) that a function $g\in \mathcal{H}_{p}(\beta,\lambda)$ if and only if

$ \frac{zg^{'}(z)+\beta z^{2}g^{''}(z)+g(z)}{zg^{'}(z)+\beta z^{2}g^{''}(z)+(2\gamma-1)g(z)} \neq \sigma\qquad(z\in \mathbb{U};\ \sigma\in\mathbb{C};\ |\sigma|=1),$

which is equivalent to

$\begin{equation}\label{3.18} \frac{(g*\mathfrak{h})(z)}{z^{-p}}\neq 0\qquad(z\in \mathbb{U}), \end{equation}$

(3.19)

where

$\begin{equation}\label{3.19} \mathfrak{h}(z)=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}c_{k}z^{k} \left(c_{k}:=\frac{k+\beta k(k-1)+1-[k+\beta k(k-1)+(2\gamma-1)]\sigma}{-p+\beta p(p+1)+1+[p-\beta p(p+1)-(2\gamma-1)]\sigma}\right). \end{equation}$

(3.20)

It follows from (3.20) that

$\begin{align*} |c_{k}| &= \left| \frac{k+\beta k(k-1)+1-[k+\beta k(k-1)+(2\gamma-1)]\sigma}{-p+\beta p(p+1)+1+[p-\beta p(p+1)-(2\gamma-1)]\sigma}\right| \\ & \leqq \frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}\qquad(|\sigma|=1). \end{align*}$

If $ f\in\Sigma_{p}$ given by (1.1) satisfies the condition (3.17), we deduce from (3.19) that

$ \frac{(f*\mathfrak{h})(z)}{z^{-p}}\neq -\varepsilon\qquad(|\varepsilon|<\delta;\ \delta>0) ,$

or equivalently,

$\begin{equation}\label{3.20} |\frac{(f*\mathfrak{h})(z)}{z^{-p}}|\geqq \delta\qquad(z\in \mathbb{U};\delta>0). \end{equation}$

(3.21)

We now suppose that

$q(z)=\frac{1}{z^{p}}+\sum\limits^{\infty}_{k=1}d_{k}z^{k}\in \mathcal{N}_{\delta}(f).$

It follows from (3.16) that

$\begin{equation}\label{3.21} \left|\frac{[(q-f)*\mathfrak{h}])(z)}{z^{-p}} \right|=\left|\sum\limits^{\infty}_{k=1}(d_{k}-a_{k})c_{k}z^{k+p}\right| \leqq |z|\sum\limits^{\infty}_{k=1}\frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}|d_{k}-a_{k}|<\delta. \end{equation}$

(3.22)

Combining (3.21) and (3.22), we easily find that

$|\frac{(q*\mathfrak{h})(z)}{z^{-p}}|=|\frac{([f+(q-f)]*\mathfrak{h})(z)}{z^{-p}}|\geqq |\frac{(f*\mathfrak{h})(z)}{z^{-p}}|-|\frac{[(q-f)*\mathfrak{h}](z)}{z^{-p}}|>0,$

which implies that

$\frac{(q*\mathfrak{h})(z)}{z^{-p}}\neq 0\qquad(z\in \mathbb{U}).$

Therefore, we have

$q(z)\in \mathcal{N}_{\delta}(f)\subset \mathcal{H}_{p}(\beta,\lambda).$

Next, we derive the partial sums of the class $\mathcal{H}_{p}(\beta,\lambda)$. For some recent investigations involving the partial sums in analytic function theory, one can refer to [3-7].

Theorem 3.4 Let $ f\in\Sigma_{p}$ be given by (1.1) and define the partial sums $f_{n}(z)$ of $f$ by

$\begin{equation}\label{3.22} f_{n}(z)=\frac{1}{z^{p}}+\sum\limits^{n}_{k=1}a_{k}z^{k} (n\in \mathbb{N}). \end{equation}$

(3.23)

If

$\begin{equation}\label{3.23} \sum\limits^{\infty}_{k=1}\frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}|a_{k}|\leqq 1, \end{equation}$

(3.24)

where $\gamma$ given by (3.3) and the condition (3.1) holds true, then

(ⅰ) $ f\in \mathcal{H}_{p}(\beta,\lambda);$

(ⅱ)

$\begin{equation}\label{3.24} \Re\left(\frac{f(z)}{f_n(z)}\right)\geqq \frac{n+\beta n(n+1)+\beta p(p+1)-p+1+2\gamma}{n+\beta n(n+1)+1+\gamma}\qquad(n\in \mathbb{N};z\in \mathbb{U}), \end{equation}$

(3.25)

and

$\begin{equation}\label{3.25} \Re\left(\frac{f_n(z)}{f(z)}\right)\geqq \frac{n+\beta n(n+1)+\beta p(p+1)-p+2+\gamma-2\beta}{n+\beta n(n+1)+2-2\beta}\qquad(n\in \mathbb{N};z\in \mathbb{U}). \end{equation}$

(3.26)

The bounds in (3.25) and (3.26) are sharp.

Proof First of all, we suppose that $f_{1}(z)=\frac{1}{z^{p}}.$ We know that

$\frac{f(z)+\varepsilon z^{-p}}{1+\varepsilon}= \frac{1}{z^{p}}\in \mathcal{H}_{p}(\beta,\lambda).$

From (3.24), we easily find that

$ \sum\limits^{\infty}_{k=1}\frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}|a_{k}-0|\leqq 1,$

which implies that $f\in N_{1}(z^{-p})$. By virtue of Theorem 3.3, we deduce that

$f\in \mathcal{N}_{1}(z^{-p})\subset \mathcal{H}_{p}(\beta,\lambda).$

Next, it is easy to see that

$\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}>\frac{n+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}>1\qquad(n\in \mathbb{N}). $

Therefore, we have

$\begin{equation}\label{3.26} \sum\limits^{n}_{k=1}|a_{k}|+\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}|a_{k}|\leqq \sum\limits^{\infty}_{k=1}\frac{k+\beta k(k-1)+\gamma}{p-\beta p(p+1)-\gamma}|a_{k}|\leqq 1. \end{equation}$

(3.27)

We now suppose that

$\begin{align}\begin{split}\label{3.27} h_{1}(z) & =\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}\left(\frac{f(z)}{f_{n}(z)}-\frac{n+\beta n(n+1)+\beta p(p+1)-p+1+2\gamma}{n+1+\beta n(n+1)+\gamma}\right) \\& =1+\frac{\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}a_{k}z^{k+p}}{1+\sum\limits^{n}_{k=1}a_{k}z^{k+p}}. \end{split}\end{align}$

(3.28)

It follows from (3.27) and (3.28) that

$\begin{equation*} |\frac{h_{1}(z)-1}{h_{1}(z)+1}|\leqq \frac{\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}|a_{k}|}{2-2\sum\limits^{n}_{k=1}|a_{k}|-\frac{n+1+\beta n(n+1)+\gamma}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}|a_{k}|}\leqq 1\qquad(z\in \mathbb{U}), \end{equation*}$

which shows that

$\begin{equation}\label{3.28} \Re(h_{1}(z))\geqq 0\qquad(z\in \mathbb{U}). \end{equation}$

(3.29)

Combining (3.28) and (3.29), we deduce that the assertion (3.25) holds true.

Furthermore, if we put

$\begin{equation}\label{3.29} f(z)= \frac{1}{z^{p}}-\frac{p-\beta p(p+1)-\gamma}{n+1+\beta n(n+1)+\gamma}z^{n+1}, \end{equation}$

(3.30)

then

$\begin{align}\begin{split} \frac{f(z)}{f_{n}(z)}&=1-\frac{p-\beta p(p+1)-\gamma}{n+1+\beta n(n+1)+\gamma}z^{n+1+p} \\&\rightarrow \frac{n+\beta n(n+1)+\beta p(p+1)-p+1+2\gamma}{n+\beta n(n+1)+1+\gamma}\qquad (z\rightarrow 1^{-}), \end{split}\end{align}$

(3.31)

which implies that the bound in (3.25) is the best possible for each $n\in \mathbb{N}.$

Similarly, we suppose that

$\begin{align}\begin{split}\label{3.30} h_{2}(z) & =\frac{n+\beta n(n+1)+2-2\beta}{p-\beta p(p+1)-\gamma}\left(\frac{f_{n}(z)}{f(z)}-\frac{n+\beta n(n+1)+\beta p(p+1)-p+\gamma+2-2\beta}{n+\beta n(n+1)+2-2\beta}\right)\\ & =1-\frac{\frac{n+\beta n(n+1)+2-2\beta}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}a_{k}z^{k+p}}{1+\sum\limits^{\infty}_{k=1}a_{k}z^{k+p}}. \end{split}\end{align}$

(3.32)

In view of (3.27) and (3.32), we conclude that

$\begin{equation*}|\frac{h_{2}(z)-1}{h_{2}(z)+1}|\leqq \frac{\frac{n+\beta n(n+1)+2-2\beta}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}|a_k|} {2-2\sum\limits^{n}_{k=1}|a_k|-\frac{n+\beta n(n+1)+2\gamma+2\beta}{p-\beta p(p+1)-\gamma}\sum\limits^{\infty}_{k=n+1}|a_k|}\leqq 1\qquad(z\in \mathbb{U}), \end{equation*}$

which implies that

$\begin{equation}\label{3.31} \Re(h_{2}(z))\geqq 0\qquad(z\in \mathbb{U}). \end{equation}$

(3.33)

Combining (3.32) and (3.33), we readily get the assertion (3.26) of Theorem 3.4. The bound in (3.26) is sharp with the extremal function $f$ given by (3.30).

Finally, we prove the following inclusion relationship for the function class $\mathcal{H}_{p}(\beta,\lambda).$

Theorem 3.5 Let $\beta_{1}\geqq \beta_{2}\geqq 1$ and $ p-\frac{1}{2}\leqq \lambda_{1}\leqq \lambda_{2}<p.$ Then

$\mathcal{H}_{p}(\beta_{1},\lambda_{1})\subset \mathcal{H}_{p}(\beta_{2},\lambda_{2}).$

Proof Suppose that $f\in \mathcal{H}_{p}(\beta_{1},\lambda_{1}).$ Then

$\Re\left(\frac{zf^{'}(z)}{f(z)}+\beta_{1}\frac{z^{2}f^{''}(z)}{f(z)}\right)<\lambda_{1}\left[\beta_{1}\left(\lambda_{1}+\frac{1}{2}\right) -1\right]+\frac{p\beta_{1}}{2}\qquad(z\in \mathbb{U}).$

Since $\beta_{1}\geqq \beta_{2}\geqq 1$ and $p-\frac{1}{2}\leqq \lambda_{1}\leqq \lambda_{2}<p,$ we find that

$\lambda_{1}\left[\beta_{1}\left(\lambda_{1}+\frac{1}{2}\right)-1\right]+\frac{p\beta_{1}}{2}\leqq \lambda_{2}\left[ \beta_{1}\left(\lambda_{2}+\frac{1}{2}\right)-1\right]+\frac{p\beta_{1}}{2}.$

Then we obtain

$\Re\left(\frac{zf^{'}(z)}{f(z)}+\beta_{1}\frac{z^{2}f^{''}(z)}{f(z)}\right)<\lambda_{2}\left[\beta_{1}\left(\lambda_{2}+\frac{1}{2}\right) -1\right]+\frac{p\beta_{1}}{2}\qquad(z\in \mathbb{U}),$

which shows that $f\in \mathcal{H}_{p}(\beta_{1},\lambda_{2}).$ By Lemma 2.2, we see that $f\in \mathcal{MS}_{p}^{*}(\lambda_{2})$, that is

$\Re\left(\frac{zf^{'}(z)}{f(z)}\right)<-\lambda_{2}\qquad(z\in \mathbb{U}).$

Now, by setting $\delta=\frac{\beta_{2}}{\beta_{1}},$ so that $0<\delta\leqq 1,$ we easily find that

$\begin{align*}\begin{split}&\Re\left(\frac{zf^{'}(z)}{f(z)}+\beta_{2}\frac{z^{2}f^{''}(z)}{f(z)}-\lambda_{2} \left[\beta_{2}\left(\lambda_{2}+\frac{1}{2}\right)-1\right]-\frac{p\beta_{2}}{2}\right) \\=&\delta \Re\left(\frac{zf^{'}(z)}{f(z)}+\beta_{1}\frac{z^{2}f^{''}(z)}{f(z)}-\lambda_{2}\left[\beta_{1}\left(\lambda_{2} +\frac{1}{2}\right)-1\right]-\frac{p\beta_{1}}{2}\right) \\&+(1-\delta)\Re\left(\frac{zf^{'}(z)}{f(z)}+\lambda_{2}\right)<0\qquad(z\in\mathbb{U}),\end{split} \end{align*}$

that is $f\in \mathcal{H}_{p}(\beta_{2},\lambda_{2})$. The proof of Theorem 3.5 is thus completed.