Lobachevskii Journal of Mathematics

http://ljm.ksu.ru

Vol. 19, 2005, 41–50 \begin{matrix}c \\ ◯ \end{matrix} Zhi-Gang Wang

Zhi-Gang Wang

A NEW SUBCLASS OF QUASI-CONVEX FUNCTIONS WITH RESPECT TO K-SYMMETRIC POINTS

(submitted by M. A. Malakhaltsev)

Abstract. In the present paper, we introduce a new subclass C_{s}^{(k)}(α, β) 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.

1. Introduction, Definitions and Preliminaries

Let A denote the class of functions of the form

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n},

which are analytic in the open unit disk

U = {z : z ∈ C and |z| < 1}.

Also let S denotes the subclass of A consisting of all functions which are univalent in U.

We denote by S^{∗}, K, C and C^{∗} the familiar subclasses of A consisting of functions which are, respectively, starlike, convex, close-to-convex and quasi-convex in U. Thus, by definition, we have (see, for details, [1] and

k-symmetric points..

2000 Mathematical Subject Classification. 30C45, 30C50.

Key words and phrases. Quasi-convex functions, subordination,

This work is supported by Scientific Research Fund of Hunan Provincial Education Department, PRC.

[2]; see also [3] and [4])

S^{∗} = \{f : f ∈ A and ℜ \{\frac{zf′(z)}{f(z)}\} > 0 (z ∈ U)\} ,

K = \{f : f ∈ A and ℜ \{1 + \frac{zf′′(z)}{f′(z)}\} > 0 (z ∈ U)\} ,

C = \{f : f ∈ A, g ∈ S^{∗}, and ℜ \{\frac{zf′(z)}{g(z)}\} > 0 (z ∈ U)\} ,

and

C^{∗} = \{f : f ∈ A, g ∈ K, and ℜ \{\frac{(zf′(z))′}{g′(z)} \} > 0 (z ∈ U)\} .

Let f(z) and F (z) be analytic in U. Then we say that the function f(z) is subordinate to F (z) in U, if there exists an analytic function ω(z) in U such that |ω(z)| ≤ |z| and f(z) = F (ω(z)), denoted by f ≺ F or f(z) ≺ F (z). If F (z) is univalent in U, then the subordination is equivalent to f(0) = F (0) and f(U) ⊂ F (U) (see [5]).

Sakaguchi [6] once introduced a class S_{s}^{∗} of functions starlike with respect to symmetric points, it consists of functions f(z) ∈ S satisfying

ℜ \{\frac{zf′(z)}{f(z) − f(−z)}\} > 0 (z ∈ U).

Following him, many authors discussed this class and its subclasses (see [7]-[14]). Motivated by S_{s}^{∗}, we can easily obtain the following class C_{s}^{∗} of functions convex with respect to symmetric points.

Definition 1. Let C_{s}^{∗} denote the class of functions in S satisfying the inequality

ℜ \{\frac{(zf′(z))′}{f′(z) + f′(−z)}\} > 0 (z ∈ U).

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 C_{s}^{(k)}(α, β) denote the class of functions in S satisfying the inequality

|\frac{(zf′(z))′}{f′_{k}(z)} − 1| < β |\frac{α(zf′(z))′}{f′_{k}(z)} + 1| (z ∈ U),  (1.1)

where 0 ≤ α ≤ 1, 0 < β ≤ 1, k ≥ 1 is a fixed positive integer and f_{k}(z) is defined by the following equality

f_{k}(z) = \frac{1}{k} ∑_{ν=0}^{k−1} ε^{−ν}f(ε^{ν}z) (ε^{k} = 1).  (1.2)

It is easy to know that C_{s}^{(2)}(1, 1) = C_{s}^{∗}, so C_{s}^{(k)}(α, β) is a generalization of C_{s}^{∗}.

In the present paper, we will discuss the integral representation and coefficient inequalities of functions belonging to the class C_{s}^{(k)}(α, β).

2. Coefficient Estimate of Functions in the Class C_{s}^{(k)}(α, β)

First we give two meaningful conclusions about the class C_{s}^{(k)}(α, β).

Theorem 1. The function f(z) ∈ C_{s}^{(k)}(α, β) if and only if

\frac{(zf′(z))′}{f′_{k}(z)} ≺ \frac{1 + βz}{1 − αβz} (z ∈ U),  (2.1)

where ” ≺ ” stands for the subordination.

Proof. Suppose that f(z) ∈ C_{s}^{(k)}(α, β), then from (1.1) we have

|\frac{(zf′(z))′}{f′_{k}(z)} − 1|^{2} < β^{2} |\frac{α(zf′(z))′}{f′_{k}(z)} + 1|^{2} .

Expanding it we get

(1 − α^{2}β^{2}) |\frac{(zf′(z))′}{f′_{k}(z)} |^{2} − 2(1 + αβ^{2})ℜ \{\frac{(zf′(z))′}{f′_{k}(z)} \} < β^{2} − 1.

If α ≠ 1 or β ≠ 1, we have

|\frac{(zf′(z))′}{f′_{k}(z)} |^{2} − \frac{2(1 + αβ^{2})}{1 − α^{2}β^{2}}ℜ \{\frac{(zf′(z))′}{f′_{k}(z)} \}

+ (\frac{1 + αβ^{2}}{1 − α^{2}β^{2}} )^{2} < \frac{β^{2} − 1}{1 − α^{2}β^{2}} + (\frac{1 + αβ^{2}}{1 − α^{2}β^{2}} )^{2} ,

that is,

|\frac{(zf′(z))′}{f′_{k}(z)} − \frac{1 + αβ^{2}}{1 − α^{2}β^{2}} |^{2} < \frac{β^{2}(1 + α)^{2}}{(1 − α^{2}β^{2})^{2}} ,

or equivalently,

|\frac{(zf′(z))′}{f′_{k}(z)} − \frac{1 + αβ^{2}}{1 − α^{2}β^{2}} | < \frac{β(1 + α)}{1 − α^{2}β^{2}}.

This tells us that the value region of G(z) = (zf′(z))′/f′_{k}(z) is contained in the disk whose center is (1 + αβ^{2})/(1 − α^{2}β^{2}) and radius is [β(1 + α)]/(1−α^{2}β^{2}). And we know that the function ω = p(z) = (1+βz)/(1− αβz) maps the unit disk to the disk:

|ω − \frac{1 + αβ^{2}}{1 − α^{2}β^{2}} | < \frac{β(1 + α)}{1 − α^{2}β^{2}}.

Notice that G(0) = p(0), G(U) ⊂ p(U), and p(z) is univalent in U, we obtain the following conclusion

\frac{(zf′(z))′}{f′_{k}(z)} ≺ p(z) = \frac{1 + βz}{1 − αβz} .

Conversely, let

\frac{(zf′(z))′}{f′_{k}(z)} ≺ \frac{1 + βz}{1 − αβz} ,

then

\frac{(zf′(z))′}{f′_{k}(z)} = \frac{1 + βω(z)}{1 − αβω(z)},  (2.2)

where ω(z) is analytic in U, and ω(0) = 0, |ω(z)| < 1. By calculation we can easily obtain from (2.2) that

|\frac{(zf′(z))′}{f′_{k}(z)} − 1| < β |\frac{α(zf′(z))′}{f′_{k}(z)} + 1| ,

that is f(z) ∈ C_{s}^{(k)}(α, β).

If α = β = 1, inequality (1.1) becomes

|\frac{(zf′(z))′}{f′_{k}(z)} − 1| < |\frac{(zf′(z))′}{f′_{k}(z)} + 1| .

It is obvious that

\frac{(zf′(z))′}{f′_{k}(z)} ≺ \frac{1 + z}{1 − z} .

Therefore, the proof of Theorem 1 is complete.

Remark 1. From Theorem 1 we know that

ℜ \{\frac{(zf′(z))′}{f′_{k}(z)} \} > 0 (z ∈ U).  (2.3)

Because of

ℜ \{\frac{1 + βz}{1 − αβz} \} > 0 (z ∈ U).

Theorem 2. Let f(z) ∈ C_{s}^{(k)}(α, β), then f_{k}(z) ∈ K ⊂ S.

Proof. For f(z) ∈ C_{s}^{(k)}(α, β), we can obtain inequality (2.3) from Theorem 1. Substituting z by ε^{μ}z in (2.3) respectively (μ = 0, 1, 2, …, k − 1; ε^{k} = 1), then (2.3) is also true, that is,

ℜ \{\frac{f′(ε^{μ}z) + ε^{μ}zf′′(ε^{μ}z)}{f′_{k}(ε^{μ}z)}\} > 0 (z ∈ U; μ = 0, 1, 2, …, k − 1). (2.4)

According to the definition of f_{k}(z) and ε^{k} = 1, we know f′_{k}(ε^{μ}z) = f′_{k}(z). Then inequality (2.4) becomes

ℜ \{\frac{f′(ε^{μ}z) + ε^{μ}zf′′(ε^{μ}z)}{f′_{k}(z)}\} > 0 (z ∈ U).  (2.5)

Let μ = 0, 1, 2, …, k − 1 in (2.5) respectively, and sum them we can get

ℜ \{\frac{∑_{μ=0}^{k−1} f′(ε^{μ}z) + z ∑_{μ=0}^{k−1} ε^{μ}f′′(ε^{μ}z)}{f′_{k}(z)}\} > 0 (z ∈ U),

or equivalently,

ℜ \{\frac{f′_{k}(z) + zf′′_{k}(z)}{f′_{k}(z)}\} > 0 (z ∈ U),

that is f_{k}(z) ∈ K ⊂ S.

Remark 2. From Theorem 2 and inequality (2.3), we know that if f(z) ∈ C_{s}^{(k)}(α, β), then f(z) is a quasi-convex function. So C_{s}^{(k)}(α, β) is a subclass of the class C^{∗} of quasi-convex functions.

In order to give the coefficient estimate of functions in the class C_{s}^{(k)}(α, β), we need the following lemma.

Lemma 1 ([14]). Let f(z) = z+∑_{n=2}^{∞} a_{n}z^{n} ∈ S, g(z) = z+∑_{n=2}^{∞} b_{n}z^{n} ∈ S, and satisfy the inequality

|\frac{zf′(z)}{g(z)} − 1| < β |\frac{αzf′(z)}{g(z)} + 1| (z ∈ U),

where 0 ≤ α ≤ 1, 0 < β ≤ 1. Then for n ≥ 2, we have

|na_{n} − b_{n}|^{2} ≤ 2(1 + αβ^{2}) ∑_{k=1}^{n−1} k |a_{k}| |b_{k}| (|a_{1}| = |b_{1}| = 1).  (2.6)

Now we give the following theorem.

Theorem 3. Let f(z) ∈ C_{s}^{(k)}(α, β) ⊂ C^{∗}, then we have (i) For l ≥ 2,

\begin{matrix}\{k(l − 1)[(l − 1)k + 1]\}^{2} |a_{(l−1)k+1}|^{2} \\ ≤ 2(1 + αβ^{2}) ∑_{i=1}^{l−1}[(i − 1)k + 1]^{3} |a_{(i−1)k+1}|^{2}  (a_{1} = 1); \end{matrix}  (2.7)

(ii) For n ≥ 2, n ≠ (l − 1)k + 1,

n^{4} |a_{n}|^{2} ≤ 2(1 + αβ^{2}) ∑_{i=1}^{[\frac{n−2}{k} +1]}[(i − 1)k + 1]^{3} |a_{(i−1)k+1}|^{2} (a_{1} = 1), (2.8)

where [\frac{n−2}{k} + 1] denotes the biggest integer ≤ \frac{n−2}{k} + 1.

Proof. It is easy to know that the condition (1.1) can write as

|\frac{z(zf′(z))′}{zf′_{k}(z)} − 1| < β |\frac{αz(zf′(z))′}{zf′_{k}(z)} + 1| (z ∈ U).

Now suppose that f(z) ∈ C_{s}^{(k)}(α, β), it is well-know that

f(z) ∈ C^{∗} ⊂ S ⟺ zf′(z) ∈ C ⊂ S.

And from theorem 2 we know

f_{k}(z) ∈ K ⊂ S ⟺ zf′_{k}(z) ∈ S^{∗} ⊂ S.

So zf′(z) and zf′_{k}(z) satisfy the condition of Lemma 1. At the same time, by the definition of f_{k}(z) we have

f_{k}(z) = \frac{1}{k} ∑_{ν=0}^{k−1} ε^{−ν}f(ε^{ν}z)

= \frac{1}{k} ∑_{ν=0}^{k−1} ε^{−ν} [ε^{ν}z + ∑_{n=2}^{∞} a_{n}(ε^{ν}z)^{n}]

= z + ∑_{l=2}^{∞} a_{(l−1)k+1}z^{(l−1)k+1}.

Using Lemma 1, let n = (l − 1)k + 1 in (2.6), we can get (2.7), if n ≠ (l − 1)k + 1, n ≥ 2, from (2.6), we can get (2.8).

3. The Integral Representation of Functions in the Class C_{s}^{(k)}(α, β)

In this section, we give the integral representation of functions in the class C_{s}^{(k)}(α, β).

Theorem 4. Let f(z) ∈ C_{s}^{(k)}(α, β), then we have

f_{k}(z) = ∫_{0}^{z} exp \{\frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}ζ} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt\} dζ,  (3.1)

where f_{k}(z) is defined by equality (1.2), ω(z) is analytic in U and ω(0) = 0, |ω(z)| < 1.

A NEW SUBCLASS OF QUASI-CONVEX FUNCTIONS 47 Proof. Let f(z) ∈ C_{s}^{(k)}(α, β), from Theorem 1 we have

\frac{(zf′(z))′}{f′_{k}(z)} = \frac{1 + βω(z)}{1 − αβω(z)},  (3.2)

where ω(z) is analytic in U and ω(0) = 0, |ω(z)| < 1. Substituting z by ε^{μ}z in (3.2) respectively (μ = 0, 1, 2, …, k − 1; ε^{k} = 1), we have

\frac{f′(ε^{μ}z) + ε^{μ}zf′′(ε^{μ}z)}{f′_{k}(ε^{μ}z)} = \frac{1 + βω(ε^{μ}z)}{1 − αβω(ε^{μ}z)} (μ = 0, 1, 2, …, k − 1). (3.3)

It is easy to know that f′_{k}(ε^{μ}z) = f′_{k}(z), sum (3.3) we can obtain

\frac{(zf′_{k}(z))′}{f′_{k}(z)} = \frac{1}{k} ∑_{μ=0}^{k−1} \frac{1 + βω(ε^{μ}z)}{1 − αβω(ε^{μ}z)},  (3.4)

from equality (3.4) we get

\frac{(zf′_{k}(z))′}{zf′_{k}(z)} − \frac{1}{z} = \frac{1}{k} ∑_{μ=0}^{k−1} \frac{(1 + α)βω(ε^{μ}z)}{z(1 − αβω(ε^{μ}z))}.  (3.5)

Integrating equality (3.5), we have

log\{f′_{k}(z)\} = \frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{z} \frac{(1 + α)βω(ε^{μ}ζ)}{ζ(1 − αβω(ε^{μ}ζ))}dζ
= \frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}z} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt,

that is,

f′_{k}(z) = exp \{\frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}z} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt\} .  (3.6)

Therefore, integrating equality (3.6) we can obtain equality (3.1). Theorem 5. Let f(z) ∈ C_{s}^{(k)}(α, β), then we have

\begin{matrix}f(z) = ∫_{0}^{z} \frac{1}{ξ} ∫_{0}^{ξ} exp \{\frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}ζ} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt\} \\ ·\frac{1 + βω(ζ)}{1 − αβω(ζ)}dζdξ, \end{matrix} (3.7)

where ω(z) is analytic in U and ω(0) = 0, |ω(z)| < 1.

Proof. Let f(z) ∈ C_{s}^{(k)}(α, β), from equalities (3.2) and (3.6) we have

\begin{matrix}(zf′(z))′ = f′_{k}(z) · \frac{1 + βω(z)}{1 − αβω(z)} \\ = exp \{\frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}z} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt\} · \frac{1 + βω(z)}{1 − αβω(z)}. \end{matrix}  (3.8)

Integrating equality (3.8) we can obtain

f′(z) = \frac{1}{z} ∫_{0}^{z} exp \{\frac{1}{k} ∑_{μ=0}^{k−1} ∫_{0}^{ε^{μ}ζ} \frac{(1 + α)βω(t)}{t(1 − αβω(t))}dt\} · \frac{1 + βω(ζ)}{1 − αβω(ζ)}dζ. (3.9)

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

4. Sufficient Condition for Functions Belonging to the Class C_{s}^{(k)}(α, β)

At last, we give the sufficient condition for functions belonging to the class C_{s}^{(k)}(α, β).

Theorem 6. Let f(z) = z+∑_{n=2}^{∞} a_{n}z^{n} be analytic in U, if for 0 ≤ α ≤ 1, 0 < β ≤ 1, we have

\begin{matrix}∑_{n=1}^{∞}(nk + 1)[(1 + αβ)(nk + 1) + β − 1] |a_{nk+1}| \\ + ∑_{n≠lk+1^{n=2}}^{∞} n^{2}(1 + αβ) |a_{n}| ≤ (1 + α)β. \end{matrix}  (4.1)

Then f(z) ∈ C_{s}^{(k)}(α, β).

Proof. Suppose that f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n}, and f_{k}(z) is defined by equality (1.2). Then for z ∈ U, we have

M = |(zf′(z))′ − f′_{k}(z)| − β |α(zf′(z))′ + f′_{k}(z)|
= |1 + ∑_{n=2}^{∞} n^{2}a_{n}z^{n−1} − 1 − ∑_{n=2}^{∞} na_{n}b_{n}z^{n−1}|

−β |α + ∑_{n=2}^{∞} αn^{2}a_{n}z^{n−1} + 1 + ∑_{n=2}^{∞} na_{n}b_{n}z^{n−1}| ,

where

b_{n} = \frac{1}{k} ∑_{ν=0}^{k−1} ε^{(n−1)ν} (ε^{k} = 1).

A NEW SUBCLASS OF QUASI-CONVEX FUNCTIONS 49

Thus, for |z| = r < 1, we have

M ≤ ∑_{n=2}^{∞} n(n − b_{n}) |a_{n}| r^{n−1}

−β [(1 + α) − ∑_{n=2}^{∞} n(αn + b_{n}) |a_{n}| r^{n−1}]

< ∑_{n=2}^{∞} n[(n − b_{n}) + β(αn + b_{n})] |a_{n}| − (1 + α)β.  (4.2)

From the definition of b_{n} we know

b_{n} = \{ \begin{matrix}1, & n = lk + 1, \\ 0, & n ≠ lk + 1. \end{matrix}  (4.3)

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

M < ∑_{n=1}^{∞}(nk + 1)[(1 + αβ)(nk + 1) + β − 1] |a_{nk+1}|

+ ∑_{n≠lk+1^{n=2}}^{∞} n^{2}(1 + αβ) |a_{n}| − (1 + α)β.

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

|\frac{(zf′(z))′}{f′_{k}(z)} − 1| < β |\frac{α(zf′(z))′}{f′_{k}(z)} + 1| (z ∈ U),

that is f(z) ∈ C_{s}^{(k)}(α, β). Therefore, the proof of theorem 6 is complete.

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50 ZHI-GANG WANG

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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