Let $\mathcal {A}$ denote the functions $f_p(z)$ of the form

$ f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}\quad(a_p>0; a_{n+p}\geqslant 0; p\in{N^*=\{1, 2, \cdots\}}), $

(1.1)

which are multivalent and analytic in the unit disc $U=\{z\in{\mathcal {C}}:|z| < 1\}.$

Here, we define the general quasi-Hadamard product of the functions $f_{p, i}$ by

$ f_{p, 1}*_{\chi_1}f_{p, 2}*_{\chi_2}f_{p, 3}*\cdots*_{\chi_{s-1}}f_{p, s}=\bigg\{\prod\limits_{i=1}^{s}a_{p, i}\bigg\}z^p- (\prod\limits_{i=1}^{s-1}\chi_i)\sum\limits_{n=1}^{\infty}\bigg\{\prod\limits_{i=1}^{s}a_{n+p, i}\bigg\}z^{n+p}, $

(1.2)

where $\chi_i$ are any nonnegative real numbers and $f_{p, i}(z)\in \mathcal {A}$ are defined as (1.1), $i=1, 2, \cdots, s.$

A function $f_p(z)$ defined by (1.1) is said to be in the class $G_p^*(a, b, \sigma)$ if and only if

$ \Bigg|\frac{\frac{zf_p'(z)}{f_p(z)}-p}{\frac{bzf_p'(z)}{f_p(z)}-ap}\Bigg| < \sigma\quad z\in{U}, $

(1.3)

where $-1\leqslant a < b\leqslant1, 0 < \sigma\leqslant1.$ Moreover, let $M_p(a, b, \sigma)$ denote the class of functions $f_p(z)$ such that $\frac{zf_p'(z)}{p}$ is in the class $G_p^*(a, b, \sigma).$

We also have the following special cases on $G_p^*(a, b, \sigma)$ and $M_p(a, b, \sigma)$:

(Ⅰ) For $a_p\equiv{1}$ in (1.1), the classes $G_p^*(a, b, \sigma)\equiv{J_p^*(a, b, \sigma)}$ and $M_p(a, b, \sigma)\equiv{C_p(a, b, \sigma)}$ were studied by Raina, Nahar [1].

(Ⅱ) For $p=1, a=-1, b=\alpha, \sigma=\beta$, the classes $G_1^*(-1, \alpha, \beta)\equiv{S_0(\alpha, \beta)}$ and $M_1(-1, \alpha, \beta)\equiv{C_0(\alpha, \beta)}$ introduced by Owa [2] are well known.

Using similar arguments as given by Raina, Nahar [1], we can easily prove the following Lemmas for functions in the classes $G_p^*(a, b, \sigma)$ and $M_p(a, b, \sigma)$:

Lemma 1.1  A function $f_p(z)$ defined by (1.1) belongs to $G_p^*(a, b, \sigma)$ if and only if

$ \sum\limits_{n=1}^{\infty}\{(1+b\sigma)n+(b-a)p\sigma\}a_{n+p}\leqslant{(b-a)p\sigma a_{p}}, $

(1.4)

where $-1\leqslant b < a\leqslant1, 0 < \sigma\leqslant1, p\in{N^*=\{1, 2, \cdots\}}.$

Lemma 1.2  A function $f_p(z)$ defined by (1.1) belongs to $M_p(a, b, \sigma)$ if and only if

$ \sum\limits_{n=1}^{\infty}(\frac{n+p}{p})\{(1+b\sigma)n+(b-a)p\sigma\}a_{n+p}\leqslant{(b-a)p\sigma a_{p}}, $

(1.5)

where $-1\leqslant b < a\leqslant1, 0 < \sigma\leqslant1, p\in{N^*=\{1, 2, \cdots\}}$.

Now, we introduce a new general class of analytic functions connected with the classes $G_p^*(a, b, \sigma)$ and $M_p(a, b, \sigma)$, which is important in the following discussion.

Definition 1.1  A function $f_p(z)$ defined by (1.1) belongs to $G_{p, c}^*(a, b, \sigma)$ if and only if

$ \sum\limits_{n=1}^{\infty}(\frac{n+p}{p})^c\{(1+b\sigma)n+(b-a)p\sigma\}a_{n+p}\leqslant{(b-a)p\sigma a_{p}}, $

(1.6)

where $-1\leqslant b < a\leqslant1, 0 < \sigma\leqslant1, p\in{N^*=\{1, 2, \cdots\}}$ and $c$ is any fixed nonnegative real number.

In fact, for every nonnegative real number $c$, the class $G_{p, c}^*(a, b, \sigma)$ is nonempty as the functions of the form

$ f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\lambda_{n+p}z^{n+p}, $

(1.7)

where $a_p>0, \lambda_{n+p}\geqslant0$ and $\sum\limits_{n=1}^{\infty}\lambda_{n+p}\leqslant1$, satisfy inequality (1.6).

We note that

(Ⅰ) For $c=0$, the class $G_{p, 0}^*(a, b, \sigma)\equiv{G_p^*(a, b, \sigma)}.$

(Ⅱ) For $c=1$, the class $G_{p, 1}^*(a, b, \sigma)\equiv{M_p(a, b, \sigma)}.$

(Ⅲ) For $p=1, a=-1, b=\alpha, \sigma=\beta$, the class $G_{1, c}^*(-1, \alpha, \beta)\equiv{S_c(\alpha, \beta)}$ was studied by Aouf [3].

(Ⅳ) For any positive integer $c$, we have the inclusion relation

$ G_{p, c}^*(a, b, \sigma)\subset{G_{p, c-1}^*(a, b, \sigma)}\subset{G_{p, c-2}^*(a, b, \sigma)} \subset{\cdots}\\\subset{G_{p, 2}^*(a, b, \sigma)}\subset{M_p(a, b, \sigma)}\subset{G_p^*(a, b, \sigma)}. $

The topology of $\mathcal {A}$ is defined to be the topology of uniform convergence on compact subsets of the unit disk $U$. Suppose that $\mathcal{X}$ is a subset of the space $\mathcal {A}$, then $f\in \mathcal{X}$ is called an extreme point of $\mathcal{X}$ if and only if $f$ can not be expressed as a proper convex combination of two distinct elements of $\mathcal{X}$. The set of all extreme points of $\mathcal{X}$ is denoted by $E\mathcal{X}.$

Furthermore, a function $f$ is called a support point of a compact $\mathcal{F}$ of $\mathcal {A}$ if $f\in{\mathcal {F}}$ and if there is a continuous linear functional $J$ on $\mathcal {A}$ such that $\mathcal{R}eJ$ is non-constant on $\mathcal {F}$ and

$ \mathcal {R}eJ(f)=\max\{\mathcal {R}eJ(g):g\in{\mathcal {F}}\}. $

We shall denote the set of all support points of $\mathcal {F}$ by supp $\mathcal {F}$.

Throughout this paper we use the notation $H\mathcal {F}$ for the closed convex hull of $\mathcal {F}.$

Lemma 1.3  (see [4]) Let $\mathcal {A}$ be a locally convex linear topological space and let $\mathcal {F}$ be a compact subset of $\mathcal {A}, $ then

(ⅰ) If $\mathcal {F}$ is non-empty, then $E\mathcal {F}$ is non-empty.

(ⅱ) $HE\mathcal {F}=H\mathcal {F}.$

(ⅲ) If $H\mathcal {F}$ is compact, then $EH\mathcal {F}\subset \mathcal {F}$.

The main object of the present work is to discuss some interesting results concerning the quasi-Hadamard product of functions belonging to the class $G_{p, c}^*(a, b, \sigma), $ which extends the earlier corresponding studies in [3, 5-10]. Also, we apply this technique in Peng Zhigang [11, 12] to obtain the extreme points and support points of some important classes with $G_{p, c}^*(a, b, \sigma)$.

Theorem2.1  Let the functions $f_{p, i}$ defined by (1.1) be in the class $M_p(a, b, \sigma)$ for every $i=1, 2, 3, \cdots, m$; $m\in{N^*}$, and let the functions $g_{p, j}$ defined by (1.1) be in the class $G^*_p(a, b, \sigma)$ for every $j=1, 2, \cdots, q, q\in{N^*}$. If $\prod\limits_{i=1}^{m+q-1}\chi_i=1$ or for any $i$, $0 < \chi_i\leqslant1$, then the quasi-Hadamard product $f_{p, 1}*_{\chi_1}f_{p, 2}*\cdots*_{\chi_{m-1}}f_{p, m}*_{\chi_m}g_{p, 1}*_{\chi_{m+1}}g_{p, 2}*\cdots*_{\chi_{m+q-1}}g_{p, q}$ belongs to the class $G_{p, 2m+q-1}^*(a, b, \sigma)\subset{M_p(a, b, \sigma)}$.

Proof  To simplify the notation, we denote by

$ \mathcal{H}_p=f_{p, 1}*_{\chi_1}f_{p, 2}*\cdots *_{\chi_{m-1}}f_{p, m}*_{\chi_m}g_{p, 1}*_{\chi_{m+1}}g_{p, 2}*\cdots*_{\chi_{m+q-1}}g_{p, q}, $

the quasi-Hadamard product of the functions $f_{p, 1}, f_{p, 2}, \cdots, f_{p, m}, g_{p, 1}, \cdots, g_{p, q}.$

Clearly,

$ \mathcal{H}_p=\bigg\{\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q}b_{p, j}\bigg\}z^p-\Big(\prod\limits_{i=1}^{m+q-1}\chi_i \Big)\sum\limits_{n=1}^{\infty} \bigg\{\prod\limits_{i=1}^{m}a_{n+p, i}\prod\limits_{j=1}^{q}b_{n+p, j}\bigg\}z^{n+p}. $

(2.1)

To prove the $\mathcal{H}_p\in{G_{p, 2m+q-1}^*}$, we need to show that

$ \begin{eqnarray*} && \sum\limits_{n=1}^{\infty}\bigg[(\frac{n+p}{p})^{2m+q-1}\{n(1+b\sigma)+(b-a)p\sigma\} \bigg\{\prod\limits_{i=1}^{m+q-1}\chi_i\prod\limits_{i=1}^{m}a_{n+p, i}\prod\limits_{j=1}^{q}b_{n+p, j}\bigg\}\bigg]\\ &\leqslant&{(b-a)p\sigma\bigg\{\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q}b_{p, j}\bigg\}.} \end{eqnarray*} $

As $f_{p, i}(z)\in{M_p(a, b, \sigma)}$, then for every $i=1, 2, \cdots, m$, we have

$ \sum\limits_{n=1}^{\infty}\Big(\frac{n+p}{p}\Big)\{(1+b\sigma)n+(b-a)p\sigma\}a_{n+p, i}\leqslant{(b-a)p\sigma a_{p, i}}. $

(2.2)

Therefore, the condition $a_{n+p, i}\geqslant0$ can make sure that

$ \Big(\frac{n+p}{p}\Big)\{(1+b\sigma)n+(b-a)p\sigma\}a_{n+p, i}\leqslant{(b-a)p\sigma a_{p, i}}, i=1, 2, \cdots, m $

(2.3)

or

$ a_{n+p, i}\leqslant{\frac{(b-a)p\sigma}{\frac{n+p}{p}[(1+b\sigma)n+(b-a)p\sigma]}a_{p, i}} $

(2.4)

for every $i=1, 2, \cdots, m.$ Also, since $-1\leqslant a < b\leqslant1, 0 < \sigma\leqslant1$, it implies

$ \frac{(b-a)p\sigma}{(1+b\sigma)n+(b-a)p\sigma}\leqslant{\Big(\frac{n+p}{p}\Big)}^{-1}, $

(2.5)

so the right side of the inequality (2.4) is not greater than $(\frac{n+p}{n})^{-2}a_{p, i}$, and we obtain

$ a_{n+p, i}\leqslant{\Big(\frac{n+p}{n}\Big)^{-2}a_{p, i}} $

(2.6)

for $i=1, 2, \cdots, m$. Similarly, for $g_{p, j}(z) \in{G_p^*(a, b, \sigma)}$, from Lemma 1.1 we have

$ \sum\limits_{n=1}^{\infty}\{(1+b\sigma)n+(b-a)p\sigma\}b_{n+p, j}\leqslant{(b-a)p\sigma b_{p, j}} $

(2.7)

for every $j=1, 2, \cdots, q.$ Furthermore, we can obtain

$ b_{n+p, j}\leqslant{\Big(\frac{n+p}{p}\Big)}^{-1}b_{p, j} $

(2.8)

for every $j=1, 2, \cdots, q.$

Using (2.6) for $i=1, 2, \cdots, m, $ (2.8) for $j=1, 2, \cdots, q-1$, (2.7) for $j=q$ and following $\prod\limits_{i=1}^{m+q-1}\chi_i=1$ or for any $i$, $0 < \chi_i\leqslant1$, we have

$ \begin{eqnarray*} &&\ \ \ \ \sum\limits_{n=1}^{\infty}\bigg[(\frac{n+p}{p})^{2m+q-1}\{n(1+b\sigma)+(b-a)p\sigma\} \bigg\{\prod\limits_{i=1}^{m+q-1}\chi_i\prod\limits_{i=1}^{m}a_{n+p, i}\prod\limits_{j=1}^{q}b_{n+p, j}\bigg\}\bigg]\\ \leqslant&&{\sum\limits_{n=1}^{\infty}\bigg[(\frac{n+p}{p})^{2m+q-1}\{n(1+b\sigma)+(b-a)p\sigma\}b_{n+p, q} \\\bigg\{(\frac{n+p}{p})^{-2m}(\frac{n+p}{p})^{-(q-1)}\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q-1}b_{p, j}\bigg\}\bigg]}\\ &&={\sum\limits_{n=1}^{\infty}\{n(1+b\sigma)+(b-a)p\sigma\}b_{n+p, q} \bigg\{\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q-1}b_{p, j}\bigg\}} \leqslant\\&&{(b-a)p\sigma b_{p, q} \bigg\{\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q-1}b_{p, j}\bigg\}}\\ &&={(b-a)p\sigma \bigg\{\prod\limits_{i=1}^{m}a_{p, i}\prod\limits_{j=1}^{q}b_{p, j}\bigg\}}, \end{eqnarray*} $

and therefore $\mathcal{H}_p\in{G_{p, 2m+q-1}^*(a, b, \sigma)}$.

Furthermore, since $G_{p, 2m+q-1}^*(a, b, \sigma)\subset{G_{p, 2m+q-2}^*(a, b, \sigma)}\subset{\cdots} \subset{G_{p, 1}^*(a, b, \sigma)}\equiv{M_p(a, b, \sigma)}$, which complete the proof of Theorem 2.1.

As $G_{p, 2m-1}^*(a, b, \sigma)\subset{G_{p, 2m-2}^*(a, b, \sigma)}\subset{\cdots} \subset{G_{p, 1}^*(a, b, \sigma)}\equiv{M_p(a, b, \sigma)}$, we can obtain the following Corollary 2.1 by setting $q=0$ in Theorem 2.1.

Corollary 2.1  Let the functions $f_{p, i}$ defined by (1.1) be in the class $M_p(a, b, \sigma)$ for every $i=1, 2, 3, \cdots, m$; $m\in{N^*}$. If $\prod\limits_{i=1}^{m-1}\chi_i=1$ or for any $i$, $0 < \chi_i\leqslant1$, then the quasi-Hadamard product $f_{p, 1}*_{\chi_1}f_{p, 2}*\cdots*_{\chi_{m-1}}f_{p, m}$ belongs to the class $G_{p, 2m-1}^*(a, b, \sigma)\subset{M_p(a, b, \sigma)}$.

As $G_{p, q-1}^*(a, b, \sigma)\subset{G_{p, q-2}^*(a, b, \sigma)}\subset{\cdots} \subset{G_{p, 1}^*(a, b, \sigma)}\equiv{M_p(a, b, \sigma)}\subset{G_p^*(a, b, \sigma)}$, we can obtain the following Corollary 2.2 by setting $m=0$ in Theorem 2.1.

Corollary 2.2  Let the functions $g_{p, j}$ defined by (1.1) be in the class $G^*_p(a, b, \sigma)$ for every $j=1, 2, \cdots, q, q\in{N^*}$. If $\prod\limits_{j=1}^{q-1}\chi_j=1$ or for any $j$, $0 < \chi_j\leqslant1$, then the quasi-Hadamard product $g_{p, 1}*_{\chi_{1}}g_{p, 2}*\cdots*_{\chi_{q-1}}g_{p, q}$ belongs to the class $G_{p, q-1}^*(a, b, \sigma)\subset{M_p(a, b, \sigma)}$.

Remark 2.1  (Ⅰ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta, \chi_i=1(i=1, 2, \cdots, m+q-1)$ in Theorem 2.1, we obtain the Aouf [3, Theorem 1] and Owa [2, Theorem 8].

(Ⅱ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta, \chi_i=1(i=1, 2, \cdots, m-1)$ in Corollary 2.1, we obtain the Aouf [3, Corollary 1] and Owa [2, Theorem 7].

(Ⅲ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta, \chi_i=1(i=1, 2, \cdots, q-1)$ in Corollary 2.2, we obtain the Aouf [3, Corollary 2] and Owa [2, Theorem 6].

(Ⅳ) Obviously, $J_p^*(a, b, \sigma)\subset{G_p^*(a, b, \sigma)}$ and $C_p(a, b, \sigma)\subset{M_p(a, b, \sigma)}$, so the corresponding results in Theorem 2.1, Corollaries 2.1, 2.2 with the classes $J_p^*(a, b, \sigma)$ and $C_p(a, b, \sigma)$ defined by Raina, Nahar [1] are all right.

Theorem 2.2  The class $G^*_{p, c}(a, b, \sigma)$ is compact subset of $\mathcal {A}.$

Proof  Montel's theorem implies that the $G^*_{p, c}(a, b, \sigma)$ contained in $\mathcal {A}$ is compact if and only if $G^*_{p, c}(a, b, \sigma)$ is closed and locally uniformly bounded (see [4, p.39]). We first assume

$ f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}\in G^*_{p, c}(a, b, \sigma), $

then (1.6) gives that

$ a_{n+p}\leqslant \frac{(b-a)p\sigma a_p}{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}, \quad n=1, 2, \cdots. $

Since $|z|=r < 1$, it follows

$ |f_p(z)|\leqslant a_p|z|^p+\sum\limits_{n=1}^{\infty}a_{n+p}|z|^{n+p}\leqslant a_pr^p+\frac{(b-a)p\sigma a_p}{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\frac{r^{1+p}}{1-r}, $

which implies that $G^*_{p, c}(a, b, \sigma)$ is locally uniformly bounded.

It remains to show that $G^*_{p, c}(a, b, \sigma)$ is sequentially closed. Suppose that a sequence $\{f_p^{(k)}(z)\}$ in $G^*_{p, c}(a, b, \sigma)$ and $\{f_p^{(k)}(z)\}\rightarrow f_p(k\rightarrow \infty)$, where

$ f_p^{(k)}(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}^{(k)}z^{n+p}. $

Weierstrass' theorem asserts that $f_p\in \mathcal {A}$ (see [4, p.38]), so we can take

$ f_p=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}, $

moreover, $a_{n+p}^{(k)}\rightarrow a_{n+p}(k\rightarrow \infty)$. We next need to consider the $f_p\in G^*_{p, c}(a, b, \sigma).$ Since $f_p^{(k)}(z)\in G^*_{p, c}(a, b, \sigma)$, (1.6) implies that

$ \sum\limits_{n=1}^{M}\frac{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}{{(b-a)p\sigma a_{p}}}a_{n+p}^{(k)}\leqslant \sum\limits_{n=1}^{\infty}\frac{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}{{(b-a)p\sigma a_{p}}}a_{n+p}^{(k)}\leqslant{1} $

for any $M\in Z^+$. Thus, as $k\rightarrow\infty$, we have

$ \sum\limits_{n=1}^{M}\frac{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}{{(b-a)p\sigma a_{p}}}a_{n+p}\leqslant 1. $

Furthermore, taking $M\rightarrow+\infty$, it gives that

$ \sum\limits_{n=1}^{\infty}\frac{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}{{(b-a)p\sigma a_{p}}}a_{n+p}\leqslant{1}. $

This completes the proof of Theorem 2.2.

Theorem 2.3  The extreme points of the class $G^*_{p, c}(a, b, \sigma)$ are given by

$ \begin{eqnarray*}&& EG^*_{p, c}(a, b, \sigma) \\ &&= \Bigg\{a_pz^p, a_pz^p-\frac{(b-a)p\sigma a_p}{(\frac{1+p}{p})^c[(1+b\sigma)+(b-a)p\sigma]}z^{1+p},\\ &&a_pz^p-\frac{(b-a)p\sigma a_p}{(\frac{2+p}{p})^c[2(1+b\sigma)+(b-a)p\sigma]}z^{2+p}, \\ && \cdots, a_pz^p-\frac{(b-a)p\sigma a_p}{(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p}, \cdots\Bigg\}, \end{eqnarray*} $

where $-1\leqslant a < b\leqslant 1, 0 < \sigma\leqslant1, n\in{N^*}$.

Proof  Using similar arguments as given by Xiong et al. [13, Theorem 2.6], we can easily obtain the extreme points on $G^*_{p, c}(a, b, \sigma).$

Theorem 2.4  The support points of the class $G^*_{p, c}(a, b, \sigma)$ are given by

$ \begin{eqnarray*}&& {\rm Supp} G^*_{p, c}(a, b, \sigma)\\ &=& \Bigg\{f_p(z)\in{G^*_{p, c}(a, b, \sigma)}:f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\phi_{n+p}z^{n+p}\Bigg\}, \end{eqnarray*} $

where $-1\leqslant a < b\leqslant 1, 0 < \sigma\leqslant1, \phi_{n+p}\geqslant0, \sum\limits_{n=1}^{\infty}\phi_{n+p}\leqslant1, n\in{N^*}$ and $\phi_{n+p}=0$ for some $n\geqslant 1.$

Proof  First, let a function

$ f_{p, 0}(z)=a_pz^p-\sum\limits_{n=1}^{\infty}\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\phi_{n+p}z^{n+p}, $

where $\sum\limits_{n=1}^{\infty}\phi_{n+p}\leqslant1, \phi_{n+p}\geqslant0, \phi_i=0$ for some $i\geqslant 1+p$. In fact, (1.7) implies that $f_{p, 0}(z)\in G^*_{p, c}(a, b, \sigma).$ Now, we need to take

$ \begin{equation*} b_{n+p}=\begin{cases} 0, &n\geqslant1, n+p\neq i, \\1, &n\geqslant1, n+p=i. \end{cases} \end{equation*} $

Obviously, we have $\lim\limits_{n\rightarrow\infty}^{-}(|b_{n+p}|)^{\frac{1}{n+p}} < 1$. Furthermore, we define a functional $J$ on $\mathcal {A}$ by

$ J(f_p(z))=\sum\limits_{n=0}^{\infty}(-a_{n+p})b_{n+p}, f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}\subset{\mathcal {A}}, \, \\ g_p(z)=b_pz^p-\sum\limits_{n=1}^{\infty}b_{n+p}z^{n+p}\subset{\mathcal {A}}. $

It is clearly that the $J$ is a continuous linear functional on $\mathcal {A}$ (see [4, p.42]). Moreover, we note that $J(f_{p, 0}(z))=-a_pb_p-\frac{(b-a)p\sigma a_p} {(\frac{i}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\phi_{i}b_i=-a_pb_p-0=-a_pb_p$. However, for any function

$ f_p(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}\in{G^*_{p, c}(a, b, \sigma)}, $

(2.9)

we can note that

$ J(f_p(z))=-a_pb_p-a_ib_i\leqslant -a_pb_p\, (i\geqslant p+1). $

So we have

$ \mathcal {R}eJ(f_{p, 0})=\max\{\mathcal {R}eJ(f_p(z)):f_p(z)\in{G^*_{p, c}(a, b, \sigma)}\} $

and $\mathcal {R}eJ(f_p(z))$ are not constant on $G^*_{p, c}(a, b, \sigma)$. Hence $f_{p, 0}$ is a support point of $G^*_{p, c}(a, b, \sigma).$

Conversely, suppose that $f_{p, 0}(z)$ is a support point of $G^*_{p, c}(a, b, \sigma)$, and $J$ is a continuous linear functional on $\mathcal {A}$. Note that $\mathcal {R}e{J}$ is also a continuous linear and is non-constant on $G^*_{p, c}(a, b, \sigma)$, consequently, we have

$ \mathcal {R}eJ(f_{p, 0})=\max\{\mathcal {R}eJ(f_p(z)):f_p(z)\in{G^*_{p, c}(a, b, \sigma)}\}. $

Let

$ \mathcal{M}=\mathcal {R}eJ(f_{p, 0}) $

and

$ \mathcal{G}_J=\{f_p(z)\in{G^*_{p, c}(a, b, \sigma)}:\, \mathcal {R}eJ(f_p(z))=\mathcal{M}\}. $

On the one hand, suppose that

$ \mathcal {R}eJ(f_{p, 1})=\mathcal {R}eJ(f_{p, 2})=\mathcal{M}, $

where $f_{p, 1}\in{G_J}, f_{p, 2}\in{G_J}, 0 < t < 1.$ Then

$ \mathcal {R}eJ[tf_{p, 1}+(1-t)f_{p, 2}]=t\mathcal {R}eJ(f_{p, 1})+(1-t)\mathcal {R}eJ(f_{p, 2})=t\mathcal{M}+(1-t)\mathcal{M}=\mathcal{M} $

and so $tf_{p, 1}+(1-t)f_{p, 2}\in{\mathcal{G}_J}, $ which gives the convexity of $\mathcal{G}_J$.

On the other hand, suppose that $\mathcal {R}eJ(f_p^{(k)}(z))=\mathcal{M}$ and $f_p^{(k)}(z)\rightarrow f_p(z)$, where $f_p^{(k)}(z)\in{\mathcal{G}_J}$. Then $\mathcal {R}eJ(f_p^{(k)}(z))\rightarrow \mathcal {R}eJ(f_p(z))$ and so $\mathcal {R}eJ(f_p(z))=\mathcal{M}$, which implies that the $\mathcal{G}_J$ is closed. Furthermore, Theorem 2.2 makes sure that the class $\mathcal{G}_J\subset G^*_{p, c}(a, b, \sigma)$ is locally uniformly bounded. Therefore, the class $\mathcal{G}_J$ is a convex compact subset of $G^*_{p, c}(a, b, \sigma)$. Thus, E $\mathcal{G}_J$ is not empty (see [Lemma 1.3]). Now, suppose that $g_{p, 0}(z)\in${E $\mathcal{G}_J$} and $g_{p, 0}(z)=tg_{p, 1}(z)+(1-t)g_{p, 2}(z)$, where $0 < t < 1, g_{p, 1}(z)\in{G^*_{p, c}(a, b, \sigma)}, g_{p, 2}(z)\in{G^*_{p, c}(a, b, \sigma)}.$ Then since

$ \mathcal {R}eJ(g_{p, 1})\leqslant \mathcal{M}, \, \mathcal {R}eJ(g_{p, 2})\leqslant \mathcal{M}, \, t\mathcal {R}eJ(g_{p, 1})+(1-t)\mathcal {R}eJ(g_{p, 2})=\mathcal {R}eJ(g_{p, 0})=\mathcal{M}, $

it follows that

$ \mathcal {R}eJ(g_{p, 1})=\mathcal {R}eJ(g_{p, 2})=\mathcal{M}, $

which implies $g_{p, 1}\in{\mathcal{G}_J}, g_{p, 2}\in{\mathcal{G}_J}.$ Again, because $g_{p, 0}\in${E $\mathcal{G}_J$}, so $g_{p, 1}=g_{p, 2}=g_{p, 0}.$ Thus $g_{p, 0}\in${E $G^*_{p, c}(a, b, \sigma)$}. This shows that E $\mathcal{G}_J\subset${E $G^*_{p, c}(a, b, \sigma)$}. Suppose

$ E\mathcal{G}_J-\{a_pz^p\}=\Bigg\{a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p}: n\in{Z_1}\Bigg\}, $

where $Z_1$ is a subset of $Z_0=\{1, 2, \cdots\}.$ We assert that $Z_1$ is a proper subset of $Z_0$. In fact, if it is not the case, then

$ E\mathcal{G}_J-\{a_pz^p\}=\Bigg\{a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p}: n\in Z_0\Bigg\}. $

Since $E\mathcal{G}_J\subset \mathcal{G}_J$, it follows that

$ \mathcal {R}eJ(a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p})=\mathcal{M} $

(2.10)

for all $n\in Z_0$. Hence,

$ \mathcal {R}eJ(a_pz^p)-\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}\mathcal {R}eJ(z^{n+p})=\mathcal{M} $

(2.11)

for all $n\in Z_0$. Let $n\rightarrow +\infty.$ Since $z^{n+p}\rightarrow 0$ in the metric of $\mathcal {A}$ and $J$ is a continuous linear functional on $\mathcal {A}, $ it follows that $\mathcal {R}eJ(z^{n+p})\rightarrow 0$. Thus, By (2.10) and (2.11) we have $\mathcal {R}eJ(a_pz^p)=\mathcal{M}$ and we also find that $\mathcal {R}eJ(z^{n+p})=0$ for all $n\in Z_0.$ Furthermore, for any $f(z)=a_pz^p-\sum\limits_{n=1}^{\infty}a_{n+p}z^{n+p}\in{G^*_{p, c}(a, b, \sigma)}, $ since $J$ is continuous on $\mathcal {A}$ and $\mathcal {R}eJ(z^{n+p})=0$ for $n\in Z_0$, it follows that

$ \mathcal {R}eJ(f_p)=\mathcal {R}eJ(a_pz^p)-\sum\limits_{n=1}^{\infty}a_{n+p}\mathcal {R}eJ(z^{n+p})=\mathcal {R}eJ(a_pz^p)=\mathcal{M}, $

which contradicts the fact that $\mathcal {R}eJ$ is not constant on $G^*_{p, c}(a, b, \sigma)$. This shows that there is an integer $i(i\geqslant 1)$ not belonging to $Z_1$. In other words,

$ a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{i+p}{p})^c[(1+b\sigma)i+(b-a)p\sigma]}z^{i+p} $

is not belonging to ${E\mathcal{G}_J}.$ Because $\mathcal{G}_J$ is a convex compact set, so $\mathcal{G}_J=HE\mathcal{G}_J$ (see [Lemma 1.3]). Following Theorem 2.3, since $f_{p, 0}(z)\in{\mathcal{G}_J}$, it gives that

$ f_{p, 0}(z)=\phi_1a_pz^p+\sum\limits_{n=1}^{\infty}\phi_{n+p}f_{n+p}(z), $

(2.12)

where $\phi_1\geqslant0, \phi_{n+p}\geqslant0$ and $\phi_1+\sum\limits_{n=1}^{\infty}\phi_{n+p}=1, f_{n+p}(z)\in{E\mathcal{G}_J}.$

Because

$ a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{i+p}{p})^c[(1+b\sigma)i+(b-a)p\sigma]}z^{i+p} $

is not belonging to ${E\mathcal{G}_J}.$ So

$ \begin{eqnarray*}f_{p, 0}(z)&&=\phi_1a_pz^p-\sum\limits_{n=1, n\neq i}^{\infty}\phi_{n+p}\Big[a_pz^p-\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p}\Big]\\ &&=a_pz^p-\sum\limits_{n=1, n\neq i}^{\infty}\phi_{n+p}\frac{(b-a)p\sigma a_p} {(\frac{n+p}{p})^c[(1+b\sigma)n+(b-a)p\sigma]}z^{n+p}.\end{eqnarray*} $

We can obtain the following Corollary 2.3 and Corollary 2.4 by setting $c=0$ and $c=1$ in Theorem 2.4, respectively.

Corollary 2.3  The support points of the class $G^*_{p}(a, b, \sigma)$ are given by

$ \begin{eqnarray*}&& {\rm Supp} G^*_{p}(a, b, \sigma)\\ &=& \Bigg\{f(z)\in{G^*_{p}(a, b, \sigma)}:f(z)=a_pz^p-\sum\limits_{n=1}^{\infty}\frac{(b-a)p\sigma a_p}{(1+b\sigma)n+(b-a)p\sigma}\phi_{n+p}z^{n+p}\Bigg\}, \end{eqnarray*} $

where $-1\leqslant a < b\leqslant 1, 0 < \sigma\leqslant1, \phi_{n+p}\geqslant0, \sum\limits_{n=1}^{\infty}\phi_{n+p}\leqslant1, n\in{N^*}$ and $\phi_{n+p}=0$ for some $n\geqslant 1.$

Corollary 2.4  The support points of the class $M_{p}(a, b, \sigma)$ are given by

$ \begin{eqnarray*}&& {\rm Supp} M_{p}(a, b, \sigma)\\ &=& \Bigg\{f(z)\in{M_{p}(a, b, \sigma)}:f(z)=a_pz^p-\sum\limits_{n=1}^{\infty}\frac{(b-a)p^2\sigma a_p} {(n+p)[(1+b\sigma)n+(b-a)p\sigma]}\phi_{n+p}z^{n+p}\Bigg\}, \end{eqnarray*} $

where $-1\leqslant a < b\leqslant 1, 0 < \sigma\leqslant1, \phi_{n+p}\geqslant0, \sum\limits_{n=1}^{\infty}\phi_{n+p}\leqslant1, n\in{N^*}$ and $\phi_{n+p}=0$ for some $n\geqslant 1.$

Remark 2.2  (Ⅰ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta$ in Theorem 2.3 and Theorem 2.4, respectively, we obtain the extreme points and support points for class $S_c(\alpha, \beta)$ defined by Aouf [3].

(Ⅱ) Putting $a_p\equiv1, c=0$ in Theorem 2.3 and Theorem 2.4, respectively, we obtain the extreme points and support points for class $J^*_p(a, b, \sigma)$ defined by Raina, Nahar [1].

(Ⅲ) Putting $a_p\equiv1, c=1$ in Theorem 2.3 and Theorem 2.4, respectively, we obtain the extreme points and support points for class $C_p(a, b, \sigma)$ defined by Raina, Nahar [1].

(Ⅳ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta, c=0$ in Theorem 2.3 and Theorem 2.4, respectively, we obtain the extreme points and support points for class $S_0(\alpha, \beta)$ defined by Owa [2].

(Ⅵ) Putting $p=1, a=-1, b=\alpha, \sigma=\beta, c=1$ in Theorem 2.3 and Theorem 2.4, respectively, we obtain the extreme points and support points for class $C_0(\alpha, \beta)$ defined by Owa [2].