Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 19, 2005, 41–50

© Zhi-Gang Wang

Abstract. In the present paper, we introduce a new subclass ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$ of quasi-convex functions with respect to $k$-symmetric points. The integral representation and several coefficient inequalities of functions belonging to this class are obtained.

 Let $\mathcal{𝒜}$ denote the class of functions of the form

which are analytic in the open unit disk

Also let $\mathcal{𝒮}$ denotes the subclass of $\mathcal{𝒜}$ consisting of all functions which are univalent in $\mathcal{𝒰}$.
 We denote by ${\mathcal{𝒮}}^{\ast }$, $\mathcal{𝒦}$, $\mathcal{𝒞}$ and ${\mathcal{𝒞}}^{\ast }$ the familiar subclasses of $\mathcal{𝒜}$ consisting of functions which are, respectively, starlike, convex, close-to-convex and quasi-convex in $\mathcal{𝒰}$. Thus, by definition, we have (see, for details, [1] and [2]; see also [3] and [4])

and

 Let $f\left(z\right)$ and $F\left(z\right)$ be analytic in $\mathcal{𝒰}$. Then we say that the function $f\left(z\right)$ is subordinate to $F\left(z\right)$ in $\mathcal{𝒰}$, if there exists an analytic function $\omega \left(z\right)$ in $\mathcal{𝒰}$ such that $\left|\omega \left(z\right)\right|\le \left|z\right|$ and $f\left(z\right)=F\left(\omega \left(z\right)\right)$, denoted by $f\prec F$ or $f\left(z\right)\prec F\left(z\right)$. If $F\left(z\right)$ is univalent in $\mathcal{𝒰}$, then the subordination is equivalent to $f\left(0\right)=F\left(0\right)$ and $f\left(\mathcal{𝒰}\right)\subset F\left(\mathcal{𝒰}\right)$ (see [5]).
 Sakaguchi [6] once introduced a class ${\mathcal{𝒮}}_s^{\ast }$ of functions starlike with respect to symmetric points, it consists of functions $f\left(z\right)\in \mathcal{𝒮}$ satisfying

Following him, many authors discussed this class and its subclasses (see [7]-[14]). Motivated by ${\mathcal{𝒮}}_s^{\ast }$, we can easily obtain the following class ${\mathcal{𝒞}}_s^{\ast }$ of functions convex with respect to symmetric points.
Definition 1. Let ${\mathcal{𝒞}}_s^{\ast }$ denote the class of functions in $\mathcal{𝒮}$ satisfying the inequality

 In the present paper, we introduce the following class of analytic functions with respect to $k$-symmetric points, and obtain some interesting results.
Definition 2. Let ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$ denote the class of functions in $\mathcal{𝒮}$ satisfying the inequality

where $0\le \alpha \le 1$, $0<\beta \le 1$, $k\ge 1$ is a fixed positive integer and $f_k\left(z\right)$ is defined by the following equality

 It is easy to know that ${\mathcal{𝒞}}_s^{\left(2\right)}\left(1,1\right)={\mathcal{𝒞}}_s^{\ast }$, so ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$ is a generalization of ${\mathcal{𝒞}}_s^{\ast }$.
 In the present paper, we will discuss the integral representation and coefficient inequalities of functions belonging to the class ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.

 First we give two meaningful conclusions about the class ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.
Theorem 1. The function $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$ if and only if

where $”\prec ”$ stands for the subordination.
Proof. Suppose that $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, then from (1.1) we have

Expanding it we get

If $\alpha \ne 1$ or $\beta \ne 1$, we have

that is,

or equivalently,

This tells us that the value region of $G\left(z\right)={\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }∕f_k^{\prime }\left(z\right)$ is contained in the disk whose center is $\left(1+\alpha {\beta }^2\right)∕\left(1-{\alpha }^2{\beta }^2\right)$ and radius is $\left[\beta \left(1+\alpha \right)\right]∕\left(1-{\alpha }^2{\beta }^2\right)$. And we know that the function $\omega =p\left(z\right)=\left(1+\beta z\right)∕\left(1-\alpha \beta z\right)$ maps the unit disk to the disk:

Notice that $G\left(0\right)=p\left(0\right)$, $G\left(\mathcal{𝒰}\right)\subset p\left(\mathcal{𝒰}\right)$, and $p\left(z\right)$ is univalent in $\mathcal{𝒰}$, we obtain the following conclusion

 Conversely, let

then

where $\omega \left(z\right)$ is analytic in $\mathcal{𝒰}$, and $\omega \left(0\right)=0$, $\left|\omega \left(z\right)\right|<1$. By calculation we can easily obtain from (2.2) that

that is $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.
 If $\alpha =\beta =1$, inequality (1.1) becomes

It is obvious that

Therefore, the proof of Theorem 1 is complete.
Remark 1. From Theorem 1 we know that

Because of

Theorem 2. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, then $f_k\left(z\right)\in \mathcal{𝒦}\subset \mathcal{𝒮}$.
Proof. For $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, we can obtain inequality (2.3) from Theorem 1. Substituting $z$ by ${\varepsilon }^{\mu }z$ in (2.3) respectively $\left(\mu =0,1,2,\dots ,k-1;{\varepsilon }^k=1\right)$, then (2.3) is also true, that is,

According to the definition of $f_k\left(z\right)$ and ${\varepsilon }^k=1$, we know $f_k^{\prime }\left({\varepsilon }^{\mu }z\right)=f_k^{\prime }\left(z\right)$. Then inequality (2.4) becomes

Let $\mu =0,1,2,\dots ,k-1$ in (2.5) respectively, and sum them we can get

or equivalently,

that is $f_k\left(z\right)\in \mathcal{𝒦}\subset \mathcal{𝒮}$.
Remark 2. From Theorem 2 and inequality (2.3), we know that if $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, then $f\left(z\right)$ is a quasi-convex function. So ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$ is a subclass of the class ${\mathcal{𝒞}}^{\ast }$ of quasi-convex functions.
 In order to give the coefficient estimate of functions in the class ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, we need the following lemma.
Lemma 1 ([14]). Let $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n\in \mathcal{𝒮}$, $g\left(z\right)=z+{\sum }_{n=2}^{\infty }{b}_n{z}^n\in \mathcal{𝒮}$, and satisfy the inequality

where $0\le \alpha \le 1$, $0<\beta \le 1$. Then for $n\ge 2$, we have

 Now we give the following theorem.
Theorem 3. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)\subset {\mathcal{𝒞}}^{\ast }$, then we have

 (i) For $l\ge 2$,

 (ii) For $n\ge 2,n\ne \left(l-1\right)k+1$,

where $\left[\frac{n-2}{k}+1\right]$ denotes the biggest integer $\le \frac{n-2}{k}+1$.
Proof. It is easy to know that the condition (1.1) can write as

Now suppose that $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, it is well-know that

And from theorem 2 we know

So $z{f}^{\prime }\left(z\right)$ and $z{f}_k^{\prime }\left(z\right)$ satisfy the condition of Lemma 1. At the same time, by the definition of $f_k\left(z\right)$ we have

Using Lemma 1, let $n=\left(l-1\right)k+1$ in (2.6), we can get (2.7), if $n\ne \left(l-1\right)k+1,n\ge 2$, from (2.6), we can get (2.8).

 In this section, we give the integral representation of functions in the class ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.
Theorem 4. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, then we have

where $f_k\left(z\right)$ is defined by equality (1.2), $\omega \left(z\right)$ is analytic in $\mathcal{𝒰}$ and $\omega \left(0\right)=0$, $\left|\omega \left(z\right)\right|<1$.
Proof. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, from Theorem 1 we have

where $\omega \left(z\right)$ is analytic in $\mathcal{𝒰}$ and $\omega \left(0\right)=0$, $\left|\omega \left(z\right)\right|<1$. Substituting $z$ by ${\varepsilon }^{\mu }z$ in (3.2) respectively $\left(\mu =0,1,2,\dots ,k-1;{\varepsilon }^k=1\right)$, we have

It is easy to know that $f_k^{\prime }\left({\varepsilon }^{\mu }z\right)=f_k^{\prime }\left(z\right)$, sum (3.3) we can obtain

from equality (3.4) we get

Integrating equality (3.5), we have

that is,

Therefore, integrating equality (3.6) we can obtain equality (3.1).
Theorem 5. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, then we have

where $\omega \left(z\right)$ is analytic in $\mathcal{𝒰}$ and $\omega \left(0\right)=0$, $\left|\omega \left(z\right)\right|<1$.
Proof. Let $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$, from equalities (3.2) and (3.6) we have

Integrating equality (3.8) we can obtain

Therefore, integrating equality (3.9) we can obtain equality (3.7).

 At last, we give the sufficient condition for functions belonging to the class ${\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.
Theorem 6. Let $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n$ be analytic in $\mathcal{𝒰}$, if for $0\le \alpha \le 1$, $0<\beta \le 1$, we have

Then $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$.
Proof. Suppose that $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n$, and $f_k\left(z\right)$ is defined by equality (1.2). Then for $z\in \mathcal{𝒰}$, we have

where

Thus, for $\left|z\right|=r<1$, we have

From the definition of $b_n$ we know

Substituting (4.3) into inequality (4.2), we get

From (4.1) we know $M<0$. Thus we have

that is $f\left(z\right)\in {\mathcal{𝒞}}_s^{\left(k\right)}\left(\alpha ,\beta \right)$. Therefore, the proof of theorem 6 is complete.

References

College of Mathematics and Computing Science, Changsha University of Science and Technology, Changsha, Hunan, 410076, People’s Republic of China

E-mail address: zhigwang@163.com

Received August 31, 2005